Learning PHYSICS for IIT JEE

Tuesday, October 23, 2007

Study guide H C Verma JEE Physics Ch. 44 X-Rays

JEE Syllabus

Characteristic and continuous X-rays, Moseley's law;
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sections of the chapter

44.1 Production of X-rays
44.2 Continuous and characteristic X-rays
44.3 Soft and Hard X-rays
44.4 Moseley's law
44.5 Bragg's law
44.6 Properties and uses of X-rays
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Study Plan

Day 1

44.1 Production of X-rays
44.2 Continuous and characteristic X-rays
44.3 Soft and Hard X-rays
Worked out examples 1 to 3

Day 2
44.4 Moseley's law
44.5 Bragg's law
44.6 Properties and uses of X-rays
WOE 3 to 6

Day 3
Exercises 1 to 10

Day 4
Exercises 11 to 20

Day5

Exercises 21 t0 27

Day 6
Objectives I and II

Day 7
Questions for short answer
Concept review
Formula review

JEE question 2007 Paper I

STATEMENT-1

If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-ray do not change.

because

STATEMENT-2

When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True

Correct choice: B
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JEE 2007 Paper II

Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-ray is

(A) λ-0 = 2mcλ^2/h
(B) λ-0 = 2h/mc
(C) λ-0 = 2m^2c^2λ^3/h^2
(D) λ-0 = λ

Correct Choice: A

Cut – off wavelength corresponds to applied accelerating potential (V)

λ cut-off = hc/eV
eV = (1/2)mv^2 = P^2/2m
and P = h/λ
eV = h^2/2mλ^2 =
λ cutoff = hc/(h^2/2mλ^2) = hc*2mλ^2/h^2 = 2mcλ^2/h
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Study guide H C Verma JEE Physics Ch. 45 SEMICONDUCTORS AND DEVICES

JEE SYLLABUS
No topics

Study Guide H C Verma JEE Physics Ch.46 THE NUCLEUS

JEE SYLLABUS

Atomic nucleus; Alpha, beta and gamma radiations; Law of radioactive decay; Decay constant; Half-life and mean life; Binding energy and its calculation; Fission and fusion processes; Energy calculation in these processes.
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Sections of the chapter

46.1 Properties of a nucleus;
46.2 Nuclear forces
46.3 Binding energy
46.4 Radioactive decay
46.5 Law of radioactive decay; Decay constant; Half-life and mean life;
46.6 Properties and uses of nuclear radiation
46.7 Binding energy and its calculation;
46.8 Fission
46.9 Uranium fission reactor
46.10 Nuclear Fusion
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Study Plan

Day 1

46.1 Properties of a nucleus;
46.2 Nuclear forces
46.3 Binding energy

Day 2
Worked out examples 1 to 5

46.4 Radioactive decay
46.5 Law of radioactive decay; Decay constant; Half-life and mean life;

Day 3
46.6 Properties and uses of nuclear radiation
46.7 Energy from the nucleus
Exercises 1 to 5

Day 4

46.8 Fission
46.9 Uranium fission reactor
WOE 6 to 10

Day 5
46.10 Nuclear Fusion
WOE 11 to 15

Day 6
Exercises 6 to 15

Day 7
Exercises 16 to 25

Day 8
Exercises 26 to 35

Day 9
Exercises 36 to 45

Day 10
Exercises 46 to 53
Questions for short answer

Day 11
Objective I

Day 12
Objective II

Day 13
Concept Review

Day 14
Formula review

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For some additional practice questions on the chapter
see
http://iit-jee-physics-ps.blogspot.com/2007/10/simple-questions-ch-46-nucleus.html

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Audiovisual lectures
Lesson 54: Nuclear Reactions
www.curriki.org/nroc/Introductory_Physics_2/lesson54/Container.html

Lesson 55: Mass-Energy Equivalence
www.curriki.org/nroc/Introductory_Physics_2/lesson55/Container.html

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JEE Question 2007 paper I

In the options given below, let E(A,B)X denote the rest mass energy of a nucleus of material X and E(n)of a neutron. The correct option is
(A)E(236,92)U > E(137,53)I + E(97,39)Y + 2E(n)
(B)E(236,92)U < E(137,53)I + E(97,39)Y + 2E(n)
(C)E(236,92)U < E(140,56)Ba + E(94,36)Kr + 2E(n)
(D)E(236,92)U = E(140,56)Ba + E(94,36)Kr + 2E(n)

Sol: Ans [A]
(236,92)U breaks into smaller atoms and gives out energy.
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