A radiation of wave length 2500 $A^{0}$ is incident on a metal plate whose work function is 3.5 eV. Then the potential required to stop the fastest photo electrons emitted by the surface is :
( $h = 6.63 \times 1 0^{- 34}$ Js and $c = 3 \times 1 0^{8}$ m/s)

Question

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Answer 1

Energy of photon $E = h ν = \frac{h c}{λ}$
Given:
$λ = 2500 \times 1 0^{- 10} m$

$E = \frac{6 . 6 3 \times 3 \times 1 0 ^{- 26}}{2 5 0 0 \times 1 0 ^{- 10}} J$

$E = \frac{0 . 0 0 7 9 5 6 \times 1 0 ^{- 16}}{1 . 6 \times 1 0 ^{- 19}} e V = 4.97 e V$

Work function of metal is $3.5 e V$ .

So maximum kinetic energy of the photoelectron after emission $K = E - W = (4.97 - 3.5) e V = 1.47 e V$
So the voltage required to stop the fastest photoelectron is $1.47 V$ .

So, the answer is option (B).

Answer 2

Energy of photon

E = h ν = \frac{h c}{λ}

Given:

λ = 2500 \times 1 0^{- 10} m

E = \frac{6 . 6 3 \times 3 \times 1 0 ^{- 26}}{2 5 0 0 \times 1 0 ^{- 10}} J

E = \frac{0 . 0 0 7 9 5 6 \times 1 0 ^{- 16}}{1 . 6 \times 1 0 ^{- 19}} e V = 4.97 e V

Work function of metal is

3.5 e V

.

So maximum kinetic energy of the photoelectron after emission

K = E - W = (4.97 - 3.5) e V = 1.47 e V

So the voltage required to stop the fastest photoelectron is

1.47 V

.

So, the answer is option (B).