## 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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