99Physics

Monostable Multivibrator

2020-03-31T03:46:00.000+05:30

Monostable Multivibrator:

Monostable multivibrator has one stable state and one quasi stable state (astable state). When an external trigger applied to the circuit, the multivibrator will jump to quasi stable state from stable state.

After the period of time it will automatically set back to the stable state, for returning to the stable state multivibrator does not require any external trigger. The time period to returning to stable state circuit is always depends on the passive elements in the circuit (resistor and capacitor values

Circuit diagram:

Circuit Operation:

When there is no external trigger to the circuit the one transistor will be in saturation state and other will be in cutoff state. Q1 is in cutoff mode and put at negative potential until the external trigger to operate, Q2 is in saturation mode.

In this case, charging of C 1 through RL1 can be seen which is shown in

Below diagram

Once the external trigger is given to the input Q1 will get turn on and when the Q1 reaches the saturation the capacitor which is connected to the collector of Q1 and base of Q2 will make transistor Q2 to turn off. This is state of turn off Q2 transistor is called astable stable or quasi state.

In this case charge capacitor discharges through R1 as shown in below

Below diagram

When capacitor charges to VCC the Q2 will turn on again and automatically Q1 is turn off. So the time period for charging of capacitor through the resistor is directly proportional to the quasi or astable state of multivibrator when a external trigger occurred (t=0.69RC).

Uses of Monostable Multivibrator:

The monostable multivibrators are used as timers, delay circuits, gated circuits etc.

Strange particles

2019-08-23T14:52:00.000+05:30

In December 1947, Rochester and Butler published the cloud

chamber photograph.

Cosmic ray particles enter from the upper left and strike the lead plate

producing a neutral particle which decayed into two charged secondaries 𝜋+ 𝑎𝑛𝑑 𝜋−

A new neutral particle with at least twice the mass of the pion was found. It is known as Kaon 𝐾0

𝐾0 → 𝜋+ + 𝜋−

In 1949, Brown and her collaborators showed the decay of charged Kaon as

𝐾+ → 𝜋+ + 𝜋+ + 𝜋−

The decay of a kaon (K+) into three pions (2 π+, 1 π−) is a process that involves both weak and strong interactions.

Weak interactions: The strange antiquark ҧ 𝑠) of the kaon transmutes into an up antiquark ത 𝑢) by the emission of a W boson; the W boson subsequently decays into a down antiquark ҧ 𝑑) and an up quark (

Strong interactions: An up quark (u) emits a gluon (g) which decays into a down quark (d) and a down antiquark ҧ 𝑑 The kaons behave in some respects like heavy pions, so

the meson family was extended to include them.

In due course, many more mesons were discovered the 𝜂(𝑒𝑡𝑎), the 𝜙(𝑃ℎ𝑖), the 𝜔(𝑂𝑚𝑒𝑔𝑎), the 𝜌(Rho), and so on.

In 1950, Anderson found one more heavier than the proton a particle called as lambda lambdaΛ

Λ → 𝑝+ + 𝜋−
It suggests that the lambda belongs to the baryon family.

Why is the proton stable?

Why proton does not decay into a positron and a

photon?

𝑝+ → 𝑒+ + 𝛾 or

𝑝+ → 𝑒+ + 𝜋0

If such reactions have to go then all atoms would disintegrate. (This reaction violates conservation of lepton number 1953)

In 1938, Stackelberg proposed a law of conservation of Baryon number to account for the stability of protons.

Baryon number 𝐴=+1for all baryons ( 𝑝and 𝑛) and A=−1 for all anti baryons

The total baryon number is conserved in any physical process.

If baryon number is conserved in all physical processes,then the proton, being the lightest baryon, should not decay

Neutron beta decay is allowed

𝑛 → 𝑝+ + 𝑒− + $\overline {\nu }$

Baryon number consideration ::(1 = 1+ 0 + 0)

And also the first reaction confirming the existence of

antiproton as

𝑝 + 𝑝 → 𝑝 + 𝑝+ 𝑝 + $\OVERLINE {\p }$

Baryon number consideration ::(1+1=1+1+1−1)

In early studies of cosmic ray showers, it was found that certain particles, which have since been identified with 𝐾(Kaon) mesons and the Σ(Sigma) and Λ0(𝐿𝑎𝑚𝑏𝑑𝑎𝑛𝑒𝑢𝑡𝑟𝑎𝑙)baryons, were produced strongly

(that is, with large cross-sections of the order of millibarns

But their lifetimes are characteristic of weak interactions: 10−10 sec. These particles were always produced in pairs

K in association with either a Σ or Λ0

All this was certainly puzzling and led to a suspicion that a new quantum number might be associated with such particles. When specific reactions, such as

𝜋− + 𝑝 → 𝐾0 + Λ0

were studied with the Λ0 and 𝐾0 subsequently decaying as

Λ0 → 𝜋− + 𝑝 and

𝐾0 → 𝜋− + 𝜋+

It was observed that the Λ0 was always produced in association with a 𝐾0 and never with just a 𝜋0.

The Λ0 was also observed to be produced in association with a 𝐾+, but not with a 𝐾-

The puzzle of associated production was clarified by Murray Gell Mann and Abraham Pais, who proposed that these particles carried a new additive quantum number, which they called strangeness, which is conserved in strong production processes but violated

in weak decays. All the ordinary mesons and baryons (as well as the photon) were assumed to be non-strange (S=0)

Thus, in any strongly associated production reaction with the initial state having no strangeness, the total strangeness of the particles in the final state must also add up to zero. From the analysis of such reactions, it was deduced that the strangeness of the 𝐾 + and 𝐾0 must be opposite to that of the Σ+, Σ0, Σ−, and Λ0.

In a pion proton interaction

𝜋− + 𝑝 → 𝐾0 + Λ0
𝜋− + 𝑝 → 𝐾+ + Σ−
𝜋− + 𝑝 → 𝐾0 + Σ0

Here the 𝐾’s carry strangeness 𝑆=+1 and Σ 𝑎𝑛𝑑 Λ carries Strangeness 𝑆=−1 and the other ordinary particles carry 𝑆=0

Strange Particles They are produced abundantly on a time scale of about 10−23sec, but they decay relatively slowly (typically about 10−10sec).

The strange particles are produced by the strong force (the same one that holds the nucleus together), but they decay by the weak force (the one that accounts for beta decay and all other neutrino processes

There exists a consistent assignment of strangeness numbers to all the hadrons (baryons and mesons) that account for the observed strong processes and “explains” why the others do not occur. (The leptons and the photon don’t experience strong forces at all, so strangeness does not apply to them.)

Willis Lamb’s Nobel Prize acceptance speech 1955

When the Nobel Prizes were first awarded in 1901, physicists knew something of just two objects which are now called “elementary particles”: the electron and proton.

A deluge of other “elementary” particles appeared after 1930; neutron,

neutrino, 𝜋meson, 𝜇meson, heavier mesons, and various hyperons. In fact, “the finder of a new elementary particle used to be rewarded by a Nobel Prize, but such a discovery now ought to be punished by a $10,000 fine”. [Source: Les Prix Nobel 1955, The Nobel Foundation, Stockholm

The Eightfold Way ( 1961-1964)

In 1961, Murray Gell Mann introduced the so-called Eightfold Way

The Eightfold Way arranged the baryons and mesons into weird geometrical patterns, according to their charge and strangeness. The eight lightest

baryons fit into a hexagonal array, with two particles at the center

THE meson OCTATE

The baryon octate

the baryon decuplet

ATOMIC MASS UNIT- amu

2019-08-21T00:39:00.000+05:30

WHAT IS ATOMIC MASS UNIT- (amu)

As we know there should be some standard(invariant) physical constant for measurement mass. an atom made up with proton, neutron, and electron.
the nucleus consists of a proton as well as neutron, this two-particle also known as nucleons which contribute (almost) the total mass of an atom.
these nucleons are basic unit of mass of an atom or simply we say atomic mass unit (amu).
When the mass is expressed in amu, it roughly reflects the sum of the number of protons and neutrons in the atomic nucleus.
It is defined precisely as 1/12 of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.
Its symbol is represented by "u" (unified atomic mass unit) or Da (Dalton), although amu may still be used.
amu is the unit that was based on oxygen-16 until 1961 when it was redefined based on carbon-12.

why is carbon-12 taken as a standard reference for measuring atomic masses? why not oxygen?

The vast majority of all carbon found on Earth is 12C. Almost 99% of all carbon on Earth is of this form. While only approximately 1% of all carbon on Earth is of the 13C isotopic form, 14C is still much rarer. Only one out of every trillion carbon atoms is 14C.

Carbon-12 is the standard while measuring the atomic masses. Because no other nuclides other than carbon-12 have exactly whole-number masses in this scale. This is due to two factors:

 A) the different mass of neutrons and protons acting to change the total mass in nuclides with proton/neutron ratios other than the 1:1 ratio of carbon-12; and 

B) an exact whole-number will not be located if there exists a loss/gain of mass to difference in mean binding energy relative to the mean binding energy for carbon-12.

 Oxygen was used as the standard for quite some time, but the results seem to be different in certain experiments. This was because oxygen-17 and oxygen-18 are also present in natural oxygen this led to 2 different tables of atomic mass. The unified scale based on carbon-12, 12C, met the physicists' need to base the scale on a pure isotope while being numerically close to the chemists' scale.

How can we prove that 1 amu is equal to $1.66\times 10^{-27} kg$?

consider a sample of C-12 and take exactly 1mole of it i.e 12g.

now we know that 1 mole of carbon sample contains $6.022\times 10^{23}$ atoms of carbon-12

$\therefore$ mass of single carbon -12 atom is $ \dfrac {12g}{6.022\times 10^{23}}$
mass of single carbon-12=1.992693457$\times 10^{-23}$g
This is the mass of a single neutral carbon-12 atom.
as $\dfrac {1}{12}$ of the mass of an unbound neutral atom of carbon-12 is 1 amu
therefore, $1amu = \dfrac {1}{12}$ mass of single carbon -12

$1amu =\dfrac {1.99269345}{12}\times 10^{-23}$ g

1amu = $0.1665077881\times 10^{-23}$g
or

1amu = $0.1665077881\times 10^{-27}$Kg

JEE Main Books | Best Books for JEE Main PHYSICS

2019-08-12T01:52:00.001+05:30

JEE Main Books | Best Books for JEE Main PHYSICS

The NCERT standard books used in CBSE board are a preferred choice for JEE.
They contain the right amount of information in an easily understandable way.
This is why you should definitely read the NCERT textbooks before you go for more advanced, difficult textbooks.
Remember, you need to clear your basics first and for that, NCERT textbooks are the best.

Once your fundamentals are strong, you can use other JEE Main Books that are of a higher level.
Below is a list of some of the best JEE Main Books besides NCERT textbooks that you should definitely consider using in order to crack JEE 2019.
These books are recommended by subject matter experts after careful research.
They are the best recommended as they explain the concepts in a way that is easy for students to understand and retain them for longer.

Best Books for JEE Main Physics

Books	Author
Concepts of Physics Vol 1 and Vol 2 VOL 1 VOL 2	H C Verma
Fundamentals of Physics	Halliday & Resnick
Understanding Physics set of 5 books Understanding Physics for JEE Main and Advanced Mechanics Part 1 Understanding Physics for JEE Main and Advanced Mechanics Part 2 Understanding Physics for JEE Main and Advanced Optics and Modern Physics Understanding Physics for JEE Main and Advanced Waves and Thermodynamics Understanding Physics for JEE Main and Advanced Electricity and Magnetism	DC Pandey
40 Years’ Chapterwise Topicwise Solved Papers (2018-1979) IIT JEE Physics	DC Pandey

GATE 2020

2019-08-11T01:39:00.000+05:30

ABOUT GATE
Graduate Aptitude Test in Engineering (GATE) is an examination conducted jointly by the Indian Institute of Science (IISc), Bangalore and the seven Indian Institutes of Technology (at Bombay, Delhi, Guwahati, Kanpur, Kharagpur, Madras and Roorkee) on behalf of the National Coordination Board (NCB)-GATE, Department of Higher Education, Ministry of Human Resource Development (MHRD), Government of India.

Qualifying in GATE is a mandatory requirement for seeking admission and/or financial assistance to: (i) Master’s programs and direct Doctoral programs in Engineering/Technology/Architecture and (ii) Doctoral programs in relevant branches of Science, in the institutions supported by the MHRD and other Government agencies. Even in some colleges and institutions, which admit students without MHRD scholarship/assistantship, the GATE qualification is mandatory. Further, many Public Sector Undertakings (PSUs) have been using the GATE score in their recruitment process.

The information in this brochure is mainly categorized into Pre-Examination (Eligibility, Application Submission, Examination Centres, etc.), Examination (Pattern, Marks/Score, etc.) and Post-Examination (Answers, Contests, Results, Scorecard, etc.) sections.

GATE 2020
The GATE 2020 will be organized by Indian Institute of Technology (IIT) Delhi in online mode on February 1, 2, 8 and 9. GATE 2020 official brochure has been released by IIT Delhi. GATE 2020 will be conducted in 25 papers including the newly added subject, Biomedical Engineering. The entrance exam is a computer based one held in two sessions per day.

Application form of GATE 2020 will be available from September 3, 2019 while admit card will be released on January 3.

GATE 2020 scores will be valid for a period of three years from the date of the release

of the result i.e., March 16. The GATE 2020 scores will be used for admission purposes in

the IITs through the Common Online Admission Portal (COAP) for IITs and CCMT for NITs. Candidates who score equal to or above the qualifying cutoff will be declared as eligible for admissions to M.Tech programmes.

The qualifying cutoff for each GATE paper will be different and will be announced in the score card by IIT Delhi.Graduate Aptitude Test in Engineering (GATE) is a national-level post-graduate entrance exam that is jointly conducted by the seven IITs (Bombay, Delhi, Kanpur, Guwahati, Kharagpur, Madras and Roorkee) and the Indian Institute of Science (IISc), Bangalore for admission to M.Tech programmes offered by government as well as many private institutes.

GATE 2020 Syllabus of Physics(PH)

GATE 2020 Eligibility

The following candidates are eligible to appear in GATE:
Bachelor’s degree holders in Engineering/Technology (4 years after 10+2 or 3 years after B.Sc./Diploma in Engineering/ Technology) and those who are in the final year of such programs.
Bachelor’s degree holders in Architecture (Five years course) and those who are in the final year of such programs.
Bachelor’s degree holders of Four-year program in Science (B.S.) and those who are in the final year of such programs.
Master’s degree holders in any branch of Science/ Mathematics/ tatistics/Computer Applications or equivalent and those who are in the final year of such programs.
Holders of Four-year Integrated Master’s degree (Post B.Sc.) in Engineering/Technology
and those who are in the second or higher year of such programs.
Holders of Five-year Integrated Master’s degree or Dual Degree in Engineering/Technology
and those who are in the fourth or higher year of such programs.
Holders of Five-year integrated M.Sc. or
Five-year integrated B.Sc./M.Sc. degree and those who are in the final year of such programs.
Candidates with qualifications obtained through examinations conducted by professional
societies recognized by UPSC/AICTE/ MHRD as equivalent to B.E./B.Tech.
Those who have completed Section A of AMIE or equivalent of such professional courses are also eligible.

Important Date

Activity	Day	Date
GATE Online Application Processing System (GOAPS) Website Opens	Tuesday	3rd sept 2019
Closing Date for Submission of (Online) Application (through Website)	Tuesday	24 sept 2019
Extended Closing Date for Submission of (Online) Application (through Website)	Tuesday	1st Oct 2019
Last Date for Requesting Change of Examination City (an additional fee will be applicable)	Friday	15th Nov 2019
Admit Card will be available in the Online Application Portal (for printing)	Friday	3rd January 2020
GATE 2020 Examination Forenoon:9:30AM to 12:30PM(Tentative) Afternoon:2:30PM to 5:30PM (Tentative)	Saturday Sunday Saturday Sunday	1st Feb 2020 2nd Feb 2020 8th Feb 2020 9th Feb 2020
Announcement of the Results in the Online Application Portal	Monday	16th March 2020

Books to prepare for IIT JAM Physics (PH) Exam – Complete List

2019-08-09T13:35:00.000+05:30

Books to prepare for IIT JAM Physics (PH) Exam – Complete List

Joint Admission Test for M.Sc is the IIT JAM exam. A large number of students appear for the JAM Physics paper each year to compete for seats in the Indian Institutes of Technology. With the aim of scoring high in JAM exam, one must have an effective preparation strategy in place. Knowing the syllabus of IIT JAM is the first step towards this goal. Next comes selecting ideal books to prepare from. We have simplified the hunt for best books for IIT JAM Physics exam for test takers by listing out the most recommended books here. Take a look at them and prepare early on!

IIT JAM Books For Physics (PH)

Whereas some of you might be attending coaching for JAM exam, others prepare on their own. The strategy for exam preparation largely is common for both groups of aspirants. As Physics is a subject that combines theory and practice in deft amount, it is important to include theoretical understanding with practice of questions for IIT JAM Physics exam preparation. These books will be a good help to you for both.

Keep in mind that your self written notes, and class notes (graduation) will also play an important role. At the time of studying theory for any subject, jot down notes in your words.

The book list compiled here contains topic wise, as well as general preparation books for JAM. While deciding which ones to purchase / study from, assess compatibility with the way author has presented topics in the book first.

Among the top priority books are solved papers of IIT JAM Physics. Whether you have started preparing early or not, at least one of these are must haves.

NAME OF BOOKS	AUTHOR
IIT-JAM Joint Admission Test for M.Sc. PHYSICS 15 Year’s Solved Papers (2005-2019) and 5 Model Papers (With Explanation) 2020	JBC press
Solved Papers & Practice Sets IIT JAM (Joint Admission Test for M. Sc. from IITs) –Physics	Arihant Publications

Subject wise, you can refer to one or more of following books for each subject.

NAME OF BOOKS	AUTHOR
Intro to quantum mechanics	DAVID GRIFFITH
Quantum Mechanics: Concepts and Applications	Nouredine Zettili
Quantum Mechanics (Schaum’s Outline Series)	Yoav Peleg, Reuven Pnini, Elyahu Zaarur, Eugene Hecht
Introduction to Solid State Physics	Charles Kittel
Elementary Solid State Physics	Omar
Solid State Physics	S. O. Pillai
Statistical Physics	F Reif
Mathematical Methods for Physics and Engineering	Riley, Hobson, Bence
Mathematical Methods for Physicists	Arfken
Classical Mechanics	Herbert Goldstein
Classical Mechanics	John R. Taylor
Introduction to Classical Mechanics: With Problems and Solutions	David Morin
Heat and Thermodynamics	Mark Zemansky
Thermodynamics Kinetic Theory and Statistical Thermodynamics	F.W. Sears, G.L. Salinger
Optics	Ghatak
Optics	by Miles V. Klein , Thomas E. Furtak
Introduction to Electrodynamics	Griffith
Electronic Devices and Circuit	Boylestad / Nashelsky
Nuclear Fission and Cluster Radioactivity: An Energy-Density Functional ApproachM.A. Hooshyar, Irwin Reichstein, F. Bary Malik	M.A. Hooshyar, Irwin Reichstein, F. Bary Malik
Nuclear Physics	V. Devanathan
Atomic and Molecular Physics	Raj Kumar

Note The JAM Physics Exam Syllabus

Following are the topics from which questions are asked; this is however a concise list. For detailed syllabus, check out JAM PHYSICS EXAM SYLLABUS

1)Mathematical Methods

2)Mechanics and General Properties of Matter

3)Oscillations, Waves and Optics

4)Electricity and Magnetism

5)Kinetic theory, Thermodynamics

6)Modern Physics

7)Solid State Physics, Devices and Electronics

Which are the best books to prepare for IIT JAM Physics?

The decision for which books to study from (for JAM Physics) can be answered on two criteria:- (1) When have you started the preparation? (2) What is your current status of knowledge of the topics? Refer to books given here to prepare for JAM Physics exam. In addition, here are some excellent tips for preparation.

Theoretical knowledge is important for JAM exam. If you have started preparations early, we recommend you to spend some time understanding the theoretical concepts. Although there are no subjective questions, but having a firm grasp of theory will go a long way in helping you tackle the paper.

Assisting knowledge with problem solving skills for IIT JAM. Education experts recommend that immediately after covering a topic, solve questions on the same. You can use other books to solve questions if the book you are currently studying from does not have many exercises for practice.

JAM Physics Exam Preparation’s key highlight is solving past year papers. At least a few months before exam, start solving previous year of IIT JAM Question paper In fact, do so in a simulated exam environment. After each test paper check answers and proceed to study further.

Understanding that JAM Physics paper is not tough, but the competition is high. It is a fact that IIT JAM Physics exam is among the toughest exams you may appear for after graduation. However level of questions is easier than JRF / UGC NET exams. Also the number of test takers compared to seats available is high, so each mark counts. For the purpose of maximizing score, focus on developing speed solving skills.

Go through list of books for IIT JAM Physics Preparation given here for your exam preparation.

If you think we should add any other book, please comment below.

Method of image for a spherical conductor

2019-07-20T19:59:00.000+05:30

Method of Images for a spherical conductor

The method of images can be applied to the case of a charge in front of a grounded spherical conductor.
The method is not as straightforward as the case of plane conductors but works equally well.
Consider a charge q kept at a distance a from the centre of the grounded sphere.
We wish to obtain an expression for the potential at a point P which is at the position (r , $ \theta$) , the potential obviously will not depend on the azimuthal angle and hence that coordinate has been suppressed.
Let the point P be at a distance $ r_{1} $ from the location of the charge q.
The image charge is located at a distance b from the centre along the line joining the centre to the charge q.
The line joining the charges and the centre is taken as the reference line with respect to which
the angle $ \theta$ is measured.

Let P be at a distance b from the image charge q'. Let q’ be the image charge.

The total potential due to charge q and q' at point P is given by
$V\left( p\right) =\dfrac {1}{4\pi \varepsilon _{0}}\left( \dfrac {q}{r_{1}}+\dfrac {q}{r_{2}}\right) .............. (1)$
now from fig,
using cosine rule we get
$ r_{1}=\sqrt {r^{2}+a^{2}-2ar\cos \theta } ..............(2) $
and
$ r_{2}=\sqrt {r^{2}+a^{2}-2br\cos \theta } .......................(3)$
put (2) and (3) in equation (1) we get

$V\left( r,\theta \right) =\dfrac {1}{4\pi \varepsilon _{0}}\left[ \dfrac {q}{ \sqrt {r^{2}+a^{2}-2ar\cos \theta }}+\dfrac {q'}{ \sqrt {r^{2}+a^{2}-2br\cos \theta }}\right] $
since sphere is connected to ground hence potential vanishes at r = R for all value of $ \theta$,
also sign of q and q' must be opposite and we must have,
$0 =\left[ \dfrac {q}{ \sqrt {R^{2}+a^{2}-2aR\cos \theta }}+\dfrac {q'}{ \sqrt {R^{2}+a^{2}-2bR\cos \theta }}\right] $
squaring on both side

$q^{2}\left( R^{2}+b^{2}-2aR\cos\theta\right) =q'^{2}\left( R^{2}+b^{2}-2bR\cos\theta\right) ......(4) $
The above equation must be equal if and only if $2q^{2}aR\cos\theta=2q'^{2}bR\cos\theta$

$\begin{aligned}\cdot \\ q'=\sqrt {\dfrac {b}{a}} q \end{aligned}$
since q and are q' are in opposite in nature, therefore we have

$\begin{aligned}\cdot \\ q'= - \sqrt {\dfrac {b}{a}} q \end{aligned} ...........(5)$
now put this in equation (4)
$\therefore R^{2}+b^{2}=\dfrac {b}{a}\left( R^{2}+a^{2}\right) $
$R^{2}+b^{2}=\dfrac {b}{a}R^{2}+ba$
$R^{2}\left( 1-\dfrac {b}{a}\right) =b\left( a-b\right) $
$R^{2}\left( \dfrac {a-b}{a}\right) =b\left( a-b\right) $
$\therefore R^{2} = ab ..................(6)$
equation (5) and (6) are two important result which used to find magnitude and distance of image charge from given charge.
now put this important result in Potential expression, i,e

$V\left( r,\theta \right) =\dfrac {1}{4\pi \varepsilon _{0}}\left[ \dfrac {q}{ \sqrt {r^{2}+a^{2}-2ar\cos \theta }}+\dfrac {q'}{ \sqrt {r^{2}+a^{2}-2br\cos \theta }}\right] $
then we get

$V\left( r,\theta \right) =\dfrac {q}{4\pi \varepsilon _{0}}\left[ \dfrac {1}{ \sqrt {r^{2}+a^{2}-2ar\cos \theta }}-\dfrac {R}{ \sqrt {r^{2}a^{2}+R^{4}-2R^{2}ar\cos \theta }}\right] $
let me know in the comment section where you stuck in the equation so that i could help out you.
than you : )

Wave motion

2019-06-10T23:52:00.000+05:30

This post only for maharastra HSC students. Other may find this stuff usefull
Wave motion
Solution of all problems:

Q 1) The equation of simple harmonic progressive wave is given by

$ y=4\times \sin\pi\times \left( \dfrac {t}{0.02}-\dfrac {x}{75}\right) $ cm.

Find the Displacement and velocity of the particle at Distance of 50 cm from the origin and at the instant 0.1 second.

Given Data:

Distance of particle from Origin = x =50 cm

time = t = 0.1

To Find Out: 1) Displacement of particle = y=?

2) velocity of particle = Vp = ?

Formula:

$ 1. y=A\sin 2\times \pi \times \left( \dfrac {t}{T}-\dfrac {x}{A}\right) .......................(1) $

$ 2.V=\omega \sqrt {A^{2}-y^{2}} ...........................................................................(2) $

Solution:

let p be the particle at a distance x from origin oscillating along y axis.

$\therefore $ Displacement of particle at an instant t = 0.1 sec is given by

$y=4\times \sin\pi\times \left( \dfrac {t}{0.02}-\dfrac {x}{75}\right) ....(Given ) .........(3) $

( In below fig, x denote the distance of particle from origin and y indicate the displacement of the particle along y axis path. blue Dotted line indicate the path of vibrating particle P)

put x= 50 & t = 0.1 sec in eq (3)

$\therefore $ equation (3) becomes

$ y=4\sin \pi \left( \dfrac {0.1}{0.02}-\dfrac {50}{75}\right) $

$ \therefore y=4\sin \pi \left( 5-\dfrac {2}{3}\right) $

$\therefore y=4\times \sin \left[ \dfrac {13\times \pi }{3}\right] $

$ \therefore y=4\sin \left( 4\pi +\dfrac {\pi }{3}\right) $

$\because \sin \left( 2n\pi +\theta \right) =\sin \theta $

$ \therefore y=4\sin \left( \dfrac {\pi }{3}\right) $

$\therefore y=4\dfrac {\sqrt {3}}{2}\\$

$ \therefore y=2\sqrt {3}\\$

$ \therefore y=2\times 1.732 $

$ \therefore y=3.464 cm .......................(4) $

now given equation of displacement

$ y=4\times \sin\pi\times \left( \dfrac {t}{0.02}-\dfrac {x}{75}\right) $

$ \therefore y=4\times \sin\2pi\times \left( \dfrac {t}{0.04}-\dfrac {x}{150}\right) $

comparing above equation with standard equation i.e equation (1)

$\therefore$ we get

$A = 4 ,T = 0.04 $

Now velocity of particle can be given by

$ V=\omega \sqrt {A^{2}-y^{2}}$

$ \therefore V=\dfrac {2\pi }{T}\left( \sqrt {A^{2}-y^{2}}\right) $ . $ \because w=\dfrac {2\pi }{T} $

$ \therefore V=\dfrac {2\times 3.142}{0.04}\sqrt {4^{2}-\left( 3\cdot 464\right) ^{2}}$

$\therefore V= 314 cm/sec $

$\therefore V=3.14 m/sec $

Q2.The frequency of a tuning fork is 256 Hz and velocity of sound in air is 350 m/s. Find the Distance covered by the wave when the fork complete 16 vibrations.

Given Data:
frequency of a tuning fork = $ n = 256 Hz $
Velocity of Sound $= V= 350 m/s $
Number of vibration $ N = 16 $
To find out :
1) Distance covered by the wave = X =?

Formula: 1 ) V = n $\lambda$
2) X = $ N \lambda$
Solution:
First of all we need to find wavelength of the wave.
$\therefore V=n\lambda$
$\therefore350=256$x$\lambda$

$\therefore \lambda =\dfrac {350}{256}$
$\therefore \lambda=1.3671875$
Now Distance covered by the wave is
X=$ N \lambda$
\Therefore X$=20 X 1.3671875
Other solution will update soon.
thanks for your patience.

What is neutrino?

2019-04-07T10:48:00.004+05:30

What is neutrino ?

A neutrino is a subatomic particle that is very similar to an electron, but has no electrical charge and a very small mass, which might even be zero. Neutrinos are one of the most abundant particles in the universe. Because they have very little interaction with matter, however, they are incredibly difficult to detect. Nuclear forces treat electrons and neutrinos identically; neither participate in the strong nuclear force, but both participate equally in the weak nuclear force. Particles with this property are termed leptons. In addition to the electron (and it's anti-particle, the positron), the charged leptons include the muon (with a mass 200 times greater than that of the electron), the tau (with mass 3,500 times greater than that of the electron) and their anti-particles.

Both the muon and the tau, like the electron, have accompanying neutrinos, which are called the muon-neutrino and tau-neutrino. The three neutrino types appear to be distinct: For instance, when muon-neutrinos interact with a target, they will always produce muons, and never taus or electrons. In particle interactions, although electrons and electron-neutrinos can be created and destroyed, the sum of the number of electrons and electron-neutrinos is conserved. This fact leads to dividing the leptons into three families, each with a charged lepton and its accompanying neutrino.

To detect neutrinos, very large and very sensitive detectors are required. Typically, a low-energy neutrino will travel through many light-years of normal matter before interacting with anything. Consequently, all terrestrial neutrino experiments rely on measuring the tiny fraction of neutrinos that interact in reasonably sized detectors. For example, in the Sudbury Neutrino Observatory, a 1000 ton heavy water solar-neutrino detector picks up about 10¹² neutrinos each second. About 30 neutrinos per day are detected.

Wolfgang Pauli first postulated the existance of the neutrino in 1930. At that time, a problem arose because it seemed that both energy and angular momentum were not conserved in beta-decay. But Pauli pointed out that if a non-interacting, neutral particle--a neutrino--were emitted, one could recover the conservation laws. The first detection of neutrinos did not occur until 1955, when Clyde Cowan and Frederick Reines recorded anti-neutrinos emitted by a nuclear reactor.

Natural sources of neutrinos include the radioactive decay of primordial elements within the earth, which generate a large flux of low-energy electron-anti-neutrinos. Calculations show that about 2 percent of the sun's energy is carried away by neutrinos produced in fusion reactions there. Supernovae too are predominantly a neutrino phenomenon, because neutrinos are the only particles that can penetrate the very dense material produced in a collapsing star; only a small fraction of the available energy is converted to light. It is possible that a large fraction of the dark matter of the universe consists of primordial, Big Bang neutrinos.

The fields related to neutrino particles and astrophysics are rich, diverse and developing rapidly. So it is impossible to try to summarize all of the activities in the field in a short note. That said, current questions attracting a large amount of experimental and theoretical effort include the following: What are the masses of the various neutrinos? How do they affect Big Bang cosmology? Do neutrinos oscillate? Or can neutrinos of one type change into another type as they travel through matter and space? Are neutrinos fundamentally distinct from their anti-particles? How do stars collapse and form supernovae? What is the role of the neutrino in cosmology?

One long-standing issue of particular interest is the so-called solar neutrino problem. This name refers to the fact that several terrestrial experiments, spanning the past three decades, have consistently observed fewer solar neutrinos than would be necessary to produce the energy emitted from the sun. One possible solution is that neutrinos oscillate--that is, the electron neutrinos created in the sun change into muon- or tau-neutrinos as they travel to the earth. Because it is much more difficult to measure low-energy muon- or tau-neutrinos, this sort of conversion would explain why we have not observed the correct number of neutrinos on Earth.

8086 Microprocessors A K Ray And K M Bhurchandi pdf

2017-09-22T01:29:00.001+05:30

ADVANCED MICROPROCESSORS AND PERIPHERALS BY AK RAY & KM BHURCHANDI .PDF FREE DOWNLOAD

Advanced Microprocessors and Peripherals by AK Ray & KM Bhurchandi.

Advanced Microprocessors And Peripherals contains a detailed analysis of the basic concepts and application of technology on microprocessors. Ajoy Kumar Ray and K. M. Bhurchandi make it easy for students to approach the complexities associated with this topic using this book.
Microprocessor technology has always been a difficult field to approach. Fortunately withAdvanced Microprocessors And Peripherals by Ajoy Kumar Ray and K M Bhurchandi, various complicated aspects have been rendered easy to understand.
The book introduces basic theory, concepts, and designs and explains it through practical application.
There are various concepts that have been explained through discussions using programming and interfacing examples such as those using assembly language. The book begins with the root concept about the processors and moves on to instruction and assembler directives. There are also references made to the architectural characteristics of the programming language and the related aspects.
There are also relevant concepts like interfacing of basic peripherals, special programmable peripheral devices, and more. The approach is simplistic and caters to every student’s needs. One can find useful information related to CRT controllers, Floppy disks, and DMA in Advanced Microprocessors And Peripherals. Microcontrollers and its management as well as architecture are also dealt with in this work.

Professor Ajoy Kumar Ray has been associated with several projects that are associated with agencies such as Defense Research & Development Organization.
Apart from this book, he has also written Fuzzy Image Processing and Applications with MATLAB.
He has also been the Principal Investigator of research in the Intel Corporation from 1997 to 2004. Ray was also the Chief Investigator for a Texas Instruments Research project for Breast carcinoma research. He has also co-authored several research texts. Currently, Ray is the Vice Chancellor at Bengal Engineering and Science University, Shibpur.
Dr. K.M Bhurchandi has studied Electronics extensively and received his education from Shri Guru Gobind Singhji College of Engineering and Technology, Maharashtra.
He completed his PhD from Nagpur University in 2004. Currently, Bhurchandi is associated with the Department of Electronics and Communication Engineering as an Assistant Professor at Shri Ramdeo Baba Kamla Nehru Engineering College. He is known for his notable contribution in the field of microprocessor technology and instrumentation. Bhurchandi has always been interested in its design and architecture and has been associated with various projects concerned in this field.

I saw many people who want to download this book as pdf but they dont get anywhere.
Guys
Download the pdf file with following link

https://drive.google.com/file/d/0B3dbCB6ixjAsQ2hJZkplQnV5QWc/view?usp=drivesdk

8086 Microprocessors: some basic programming

2017-09-12T17:13:00.002+05:30

hii everyone., here we have some basic programming of 8086 microprocessors

1.Addition of two hexadecimal nos. (BYTE or WORD)
MOV AX, 1ST DATA Move 1st data byte / word in Acc reg.
MOV BX, 2ND DATA Move 2nd data byte/ word in BX reg.
ADD AX, BX Add contents of Ax and BX.
INT 21 Return to DOS

Enter the program by using the a ASSEMBLE command.

First type a 0100 <ENTER> and line by line enter the program.

Now in order to verify the program use the u UNASSEMBLE
command.
Type U 0100 <ENTER>
And debug will then display the entire program.

Since this program is entered from offset 0100, hence check
that the IP register should be at 0100. This can be done by
using the R REGISTER command. For that

Type r IP <ENTER>

The screen will show IP :0100 _

If the address of the IP register is the same as that of the
program you want to execute, then just press <ENTER> key.
But if it is different then at that blank space just after the IP
address is displayed, type the new offset address and press
<ENTER> key.
Then execute the program by using the g GO command.
Type g 0108 <ENTER> (since this program ends at 0108)
Examine the contents of the registers as well as the Flags.

Write and execute programs considering hexadecimal nos. for:
Subtraction
Multiplication
Division

Examine the contents of the registers as well as the Flags

Write and execute programs considering decimal nos. for:
i) Addition
ii) Subtraction

Examine the contents of the registers as well as the Flags

Write an assembly language program to add the contents of the memory location 1000:0100H to the contents of 1500:0100H and store the result in memory location 2500:0100H

MOV CX, 1000H Move 1000h to CX to Initialize DS
MOV DS, CX Initialize DS at 1000H
MOV AX, [0100] Move data at offset 0100 to AX
MOV CX, 1500H Move 1500h to CX to Initialize DS
MOV DS, CX Initialize DS at 1500H
MOV BX, [0100] Move data at offset 0100 to BX
ADD AX, BX Add the two data bytes.
MOV CX, 2500H Move 2500h to CX to Initialize DS
MOV DS, CX Initialize DS at 2500H
MOV [0100], AX Store the result at 2500:0100H
INT 21 Return to DOS
First type a 0200 <ENTER> and line by line enter the above
program.
Now in order to load the data in 1000:0100 & 1500:0100 use
the e command i.e. the ENTER command

Type e 1000:0100 <ENTER>
The screen will display 1000:0100 D2._
D2 is the data present earlier, now at that blank space enter
the new data byte, and then press <ENTER>.

Similarly to enter data at 1500:0100

Type e 1500:0100 <ENTER>
The screen will display 1500:0100 40._
40 is the data present earlier, now at that blank space enter
the new data byte, and then press <ENTER>.

Execute the program by using the GO command and check
2500:0100 for the result.

Ten data bytes are stored in consecutive memory locations
having offset from 0200H in the segment starting from1000H.
Write an assembly language program to move the entire block
of data bytes to a new offset address starting from 0300H.
(i.e. move data bytes stored from 1000:0200 1000:0209 to
new memory locations 1000:0300 1000:0309)
In order to understand and to develop the logic you should first write and execute this program in a general way i.e. using simple instructions.
Then try to minimize the above program by using the LOOP instruction, and
Further modify and minimize it by using the MOVSB ( move string byte ) instruction.

In order to store the data bytes use the e i.e. the ENTER command. Please note that only one data byte can be entered at a time.

Five data bytes are stored in consecutive memory locations
having offset from 0200H in the segment starting from1000H.
Write an assembly language program to find the greatest of
these five data bytes . Store the greatest number in AL register.

Write an assembly language program to find out the number of
even and odd numbers from a given ten, words i.e. 16-bit
hexadecimal nos. (The words are stored from 0300 offset
onwards)(Please note that each word will occupy two memory
locations).

Write an assembly language program to find out the number of
positive and negative numbers from a given ten, signed nos.
(The signed nos. are stored from offset 0300 onwards).

9. Write an assembly language program to display a string of
characters on the screen. Use the ASCII code chart.

10. Write an assembly language program to find the square root of
a decimal number. The number considered should be a perfect
square.

8085 program of Trapazoidal wave

2017-09-12T16:22:00.000+05:30

Hii everyone , here you will get 8085 program of Trapazoidal wave

TRAPAZOIDAL
Address	Op code	Lable	Mnemonics
F000	3E,89		MVI A,89
F002	D3,13		OUT 13
F004	3E,01		MVI A,01
F006	D3,11		OUT 11
F008	3E,00	START	MVI A,00
F00A	D3,10	LOOP1	OUT 10
F00C	3C		INR A
F00D	FE,FF		CPI FF
F00F	C2,0A,F0		JNZ F00A (LOOP1)
F012	CD,00,F1		CALL DELAY F100
F015	3E,FF		MVI A,FF
F017	D3,10	LOOP2	OUT 10
F019	3D		DCR A
F01A	FE,00		CPI 00
F01C	C2,17,F0		JNZ F017 (LOOP2)
F01F	CD,00,F1		CALL DELAY F100
F022	C3,08,F0		JMP START F008

DELAY ROUTINE
F100	16,FF	DELAY	MVI D,FF
F102	15	LOOP1	DCR D
F103	C2,02,F1		JNZ D (LOOP1)
F106	C9		RET

OUTPUT:

AMPLITUDE=

CAL. TIME PERIOD=

OBS. TIME PERIOD=

8085 traiangular wave program

2017-09-12T16:10:00.002+05:30

Hii everyone, here you will get traingular wave programme of 8085

TRIANGULAR
Address	Op code	Lable	Mnemonics
F000	3E,89		MVI A,89
F002	D3,13		OUT 13
F004	3E,01		MVI A,01
F006	D3,11		OUT 11
F008	3E,00	START	MVI A,00
F00A	D3,10	LOOP1	OUT 10
F00C	3C		INR A
F00D	FE,FF		CPI FF
F00F	C2,0A,F0		JNZ F00A (LOOP1)
F012	D3,10	LOOP2	OUT 10
F014	3D		DCR A
F015	FE,00		CPI 00
F017	C2,12,F0		JNZ F012 (LOOP2)
F01A	C3,08,F0		JMP (START)

OUTPUT:

AMPLITUDE=

CAL. TIME PERIOD=

OBS. TIME PERIOD=

Square wave 8085 program

2017-09-12T16:06:00.001+05:30

Hii everyone , here you will get square wave program of 8085

SQUARE
Address	Op code	Lable	Mnemonics
F000	3E,89		MVI A,89
F002	D3,13		OUT 13
F004	3E,01		MVI A,01
F006	D3,11		OUT 11
F008	3E,FF	LOOP	MVI A,FF
F00A	D3,10		OUT 10
F00C	CD,00,F1		CALL DELAY
F00F	3E,00		MVI A,00
F011	D3,10		OUT 10
F013	CD,00,F1		CALL DELAY F100
F016	C3,08,F0		JMP F008 (LOOP)

DELAY ROUTINE
F100	16,FF	DELAY	MVI D,FF
F102	15	LOOP1	DCR D
F103	C2,02,F1		JNZ D (LOOP1)
F106	C9		RET

OUTPUT:

AMPLITUDE=

CAL. TIME PERIOD=

OBS. TIME PERIOD=

Debug of 8086

2017-09-12T00:43:00.000+05:30

ABOUT DEBUG

DEBUG.COM is a DOS utility that facilitates the debugging and trouble-shooting of assembly language programs.

By using Debug to run a program one instruction at a time, and watching how the program works, we can find mistakes and correct them. This is known as debugging, hence the name DEBUG.

Actually, the DEBUG enables you to use the personal computer as a low-level microprocessor kit.

How to use debug

First go to MS DOS either by clicking MS-DOS prompt or command prompt.
In that drive just type Debug and press enter key. Your monitor screen will go blank and the cursor will keep on blinking waiting for your command.

Now type r <ENTER>

You will able to see all the registers, pointers, flags etc as shown below:

AX=0000 BX=0000 CX=0000 DX=0000 SP=FFEE BP=0000

SI=0000 DI=0000 DS=0D22 ES=0000 SS=0D22 CS=0D22

IP=0100 NV UP EI NG NZ NA PE CY

Where AX, BX, CX, DX are general-purpose registers.
CS, DS, SS, ES are segment registers.
SP, BP, SI, DI are pointers.

The flags are listed below with their codes for SET and RESET

FLAG NAME
SET
RESET

Carry
CY
NC

Parity
PE ( Even )
PO ( Odd )

Auxillary Carry
AC
NA

Zero
ZR
NZ

Sign
NG ( Negative )
PL ( Plus )

Interrupt
EI ( Enabled )
DI ( Disabled )

Direction
DN ( Decrement)
UP

Overflow
OV
NV

Now with help of DEBUG commands write and execute various programs.
The DEBUG commands are given from the next page onwards.

DEBUG COMMANDS

COMMAND FUNCTIONS

1) A (offset address) - ASSEMBLE Assembles mnemonics directly into the
memory.
In order to enter the program (mnemonics) in memory this command is used.

Example: Suppose that you want to write the program at the offset address of 0100
then type a 0100 <ENTER>

The screen will then display 0f64 : 0100
Where 0f64 is the starting address of the Code Segment and 0100 is the Offset
address.

Now you can directly type the instructions line by line. And after each instruction
press enter key.

Example: If you entered a 10 line program by the assemble command and now
if you want to change the instruction at 0108 , then type
a 0108 <ENTER>

The screen will then display 0f64 : 0108
Now you can change the instruction and then press the enter key.

2) C range address - COMPARE Compares the portion of memory
specified by range to a portion of the
same size beginning at the specified
address.

Example: If you type the command C 100, 1FF 300

It means that, this compares the block of memory from 100 to 1FFH with the block
of memory from 300 to 3FFH.

3) D [range] - DUMP Displays the hex code or machine code
present in that address segment.

Example: If you type D 100 <ENTER>

Then the debug displays the dump in the following format
0D22 : 0100 42 45 23 56 24 09 ..
0D22 : 010A 35 78 56 55

4) E (offset address) - ENTER Enters hex code or machine code into
the memory the specified offset
address.
Only one hex code can be entered at a time.

Example: If you type the command E 0200 <ENTER>
The screen will display
0D22 : 0200 42._
It means that, at offset address 0200 42H the hex code is already stored. If you
want to change the hex code type the new code and then press enter.

5) F range list - FILL Fills the addresses in the specified range
with the values in the specified list.

Example: If you type the command f 0D22 : 0100 L 0100 20 D8 00 01 <ENTER>

The debug would fill memory locations 0D22 : 0100 through 0D22 : 01FF with
the bytes specified. The four values would then be repeated until all the 100H
bytes were filled.

6) G [offset address] - GO Executes the program from the current
CS:IP address.

Example: If you type the command G <ENTER>

The debug would then execute the program from the current CS: IP address.

Example: If IP=0100, and you type the command G 0120 <ENTER>

The debug would then execute the program from the offset address 0100 till 0120.

7) H value value - HEX Performs hexadecimal arithmetic on the
two specified parameters.

Example: If you type the command H 019F 010A <ENTER>

The debug will perform the calculations and then display the following result
02A9 0095

The debug first adds the two parameters and then subtracts the second from the
first.

8) I (port address) INPUT Inputs and displays the data byte from the
port specified by port address.

Example: If you type the command I 02F8 <ENTER>

Suppose that the byte at the port 02F8H is 42H. Debug would input the byte and
then display the following : 42 .

9) L [add[drive: record record ]] LOAD Loads a file into the memory.

10) M range address MOVE Moves the block of memory specified
by the range to the location beginning
at the specified address.

Example: If you type the command M 0D22 :0100 0110 0D22 : 0200 <ENTER>
Debug would will move the data from 0D22 : 0100 till 0D22 : 0110 to
0D22 : 0200 till 0D22 : 0210

i.e. data bytes from 0100 0110 will be moved ( copied ) to 0200 0210 in the
same segment.

11) N filename ( filename ) NAME Sets the filenames.

12) O port address byte OUTPUT Sends the specified byte to the output at
the specified port address.

Example: If you type the command O 02F8 AF <ENTER>

It means that the data AF will be outputted to the port 02F8.

13) P [=address][number] PROCEED Executes a loop, a repeated string
instruction, a software interrupt , or a
subroutine call to completion.

The P (Proceed) command transfers control from debug to the target program.
The program executes without interruption until the loop, a repeated string
instruction, software interrupt, or sub-routine call at address is completed , or
until the specified number of machine instructions has been executed.

14) Q QUIT To exit from DEBUG and return to DOS.

Example: If you type the command q <ENTER>

You will exit Debug and return to DOS.

15) R [register name] R Displays the contents of all the registers
(general purpose, segment, pointers, flags)

Example: If you type the command R <ENTER>

The debug will then display the following
AX=0000 BX=0000 CX=0000 DX=0000 SP=FFEE BP=0000
SI=0000 DI=0000 DS=0D22 ES=0000 SS=0D22 CS=0D22
IP=0100 NV UP EI NG NZ NA PE CY

Where AX, BX, CX, DX are general-purpose registers.
CS, DS, SS, ES are segment registers.
SP, BP, SI, DI are pointers.

The flags are listed below with their codes for SET and RESET

FLAG NAME
SET
RESET

Carry
CY
NC

Parity
PE ( Even )
PO ( Odd )

Auxillary Carry
AC
NA

Zero
ZR
NZ

Sign
NG ( Negative )
PL ( Plus )

Interrupt
EI ( Enabled )
DI ( Disabled )

Direction
DN ( Decrement)
UP

Overflow
OV
NV

Example: If you to change the data in one of the registers for e.g . BX
then type R followed by the register name you want to change
i.e. R BX <ENTER>

The screen will display BX=0000.

Now type the new data and press enter key.

Example: If you type F as the register name after the R command i.e.
R F <ENTER>

Debug will display each flag with a two character alphabetic code as shown
above.

To change any flag type the opposite two-letter code. The flags are then either
SET or RESET.

16) S range list SEARCH Searches the specified range for the
specified list of data byte.

Example: If you want to search or locate the data 29H from offset 0100 to 0110
Then type the command S 0D22 : 0100 0110 29 <ENTER>

The display will show the addresses, at which the data 29h is present,
for example:
0D22:0100
0D22:0105

17) T[=address][value] TRACE Executes one instruction and displays the
contents of all registers, flags and the
decoded instruction.

This command is generally used for single stepping.

Example: If you type the command T =011A 10 <ENTER>

Debug executes sixteen instructions starting from 011A in the current segment
and then displays all registers and flags for each instruction as it is executed.
The display scrolls away until the last instruction is executed and then stops.

18) U [range] UNASSEMBLE It displays the hex code along with their
mnemonics.

Example: If you type the command U 04BA : 0100 L 09 <ENTER>

Debug would disassemble 9 bytes beginning at address 04BA : 0100
04BA : 0100 206472 AND [SI + 72 } , AH
04BA : 0103 69 DB 69
04BA : 0104 7665 JBE 016B and so on

19) W[address[drive:record record]] WRITE Writes the file being debugged to
a disk file.

ORBIT and ORBITALS

2017-09-11T16:25:00.000+05:30

welcome folks

here i have concern with the orbit and orbitals

according to Quantum mechanics we are not dealing with the actual path travel by the electrons. we always imagine there is imaginary orbit around the nucleus in which electron is orbiting. wait a minute,i just discussed the orbit.
now what is orbit, orbit is imaginary path in which probability of finding electron is very less. so where will be electron so that i can find with maximum probability. the answer is orbitals.where you could find electron with maximum probability.
currently we have four orbitals i.e s,p,d,&f orbitals and we have found out their shape experimently. there could exist g,h,.and so on.... orbitals but we don't have that element for which electron could go into g,h ... orbitals and carrying the information of that perticular orbital. therefore we dont have shape of those orbitals yet.