JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Modern Physics

The stopping potential

V_{0}

(in volt) as a function of frequency (

v

) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be :
(Given : Planck's constant

(h) = 6.63 \times 1 0^{- 34} Js

, electron charge

e = 1.6 \times 1 0^{- 19} C

)

(A)1.82 eV
(B)1.66 eV✓ Correct
(C)2.12 eV
(D)1.95 eV

Answer: (B)

Solution

Einstein's photoelectric relation

h ν = ϕ + e V_{0}

rearranges to

V_{0} = \frac{h}{e} ν - \frac{ϕ}{e}

, so the stopping-potential line meets the frequency axis at the threshold

ν_{0}

, where

ϕ = h ν_{0}

. The graph gives

ν_{0} = 4 \times 1 0^{14} Hz

, so

ϕ = 6.63 \times 1 0^{- 34} \times 4 \times 1 0^{14} = 2.652 \times 1 0^{- 19} J

. Dividing by

e

ϕ = \frac{2.652 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} \approx 1.66 eV

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Modern Physics

Solution

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