# The maximum KE of the photoelectron is found to be $6.63\times10^{-19}J$, when the metal is irradiated with a radiation of frequency $2\times10^{15} H_{2}$ . The threshold frequency of the metal is about Option 1) $1\times10^{15}s^{-1}$ Option 2) $2\times10^{15}s^{-1}$ Option 3) $3\times10^{15}s^{-1}$ Option 4) $1.5\times10^{15}s^{-1}$

P Plabita

As learnt in

Photoelectric Effect -

$\frac{1}{2}mu^{2}= hv-hv_{0}$

- wherein

where

m is the mass of the electron

u is the velocity associated with the ejected electron.

h is plank’s constant.

v is frequency of photon,

v0 is threshold frequency of metal.

$\frac{1}{2}mu^{2} =$ maximum KE

$V_{0} =$ threshold frequency

$\therefore$  we have

$6.63\times 10^{-19} = 6.63\times 10^{-34} (2\times 10^{15}hz-v_{0})$

$V_{0} = 1\times 10^{15} s^{-1}$

Option 1)

$1\times10^{15}s^{-1}$

This solution is correct.

Option 2)

$2\times10^{15}s^{-1}$

This solution is incorrect.

Option 3)

$3\times10^{15}s^{-1}$

This solution is incorrect.

Option 4)

$1.5\times10^{15}s^{-1}$

This solution is incorrect.

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