The magnetic field associated with a light wave is given, at the origin, by B = B _ 0 [ (3.14 × 10 ^ 7 )ct + (6.28 × 10 ^ 7 )ct] . If this light falls on a silver plate having a work function of 4 .7 eV, what will be

Correct answer: B. 7 .72 eV B = B _ 0 [ (3.14 × 10 ^ 7 ct) + (6.28 × 10 ^ 7 ct)] Maximum KE will associate with the maximum frequency f _ max = 6.28 × 10 ^ 7 c 2π = 10 ^ 7 × 3 × 10 ^ 8 Hz = 3 × 10 ^ 15 Hz K E _ = h f _ - ϕ = [ 6 .63 × 10 ^ - 34 × 3 × 10 ^ 15 1 .6 × 10 ^ - 19 - 4 .7 eV] = 7 .72 eV

JEE Main 20199 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Evening Shift
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2019 · 9 January · Evening Shift · Physics · Dual Nature of Matter

The magnetic field associated with a light wave is given, at the...

The magnetic field associated with a light wave is given, at the origin, by

B = {B}_{0}\left[\sin \left(3.14 \times {10}^{7}\right)ct + \sin \left(6.28 \times {10}^{7}\right)ct\right] .

If this light falls on a silver plate having a work function of

4 .7 eV,

what will be the maximum kinetic energy of the photoelectrons?

( c = 3 \times {10}^{8}m{s}^{- 1}, h = 6 .6 \times {10}^{- 34}Js )

Options

A $6 .82 eV$
B $7 .72 eV$
C $12 .5 eV$
D $8 .52 eV$

Correct answer

B. $7 .72 eV$

Step-by-step solution

B = {B}_{0}\left[\sin \left(3.14 \times {10}^{7}ct\right) + \sin \left(6.28 \times {10}^{7}ct\right)\right]

Maximum

KE

will associate with the maximum frequency

{f}_{max } = \dfrac{6.28 \times {10}^{7}c}{2\pi} = {10}^{7} \times 3 \times {10}^{8}Hz = 3 \times {10}^{15}Hz

K{E}_\max = h{f}_\max - ϕ

= \left[\dfrac{6 .63 \times {10}^{- 34} \times 3 \times {10}^{15}}{1 .6 \times {10}^{- 19}} - 4 .7 eV\right]

= 7 .72 eV

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