A metal surface of work function $1.07 e V$ is irradiated with light of wavelength $332 n m .$ The retarding potential required to stop the escape of photon electron is:

4.81 V

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3.74 V

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2.66 V

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1.07 V

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Solution

The correct option is C $2.66 V$
Energy of incident light,
$E (e V) = \frac{12375}{3320} = 3.72 e V$
relation, $E = W_{0} + e V_{0}$
now, $V_{0} = \frac{(E - W_{0})}{e} = \frac{3.72 e V - 1.07 e V}{e}$
$V_{0} = 2.65 V o l t$

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A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photon electron is:

The correct option is C 2.66 VEnergy of incident light,E(eV)=123753320=3.72eVrelation, E=W0+eV0now, V0=(E−W0)e=3.72eV−1.07eVeV0=2.65Volt

A metal surface of work function $1.07 e V$ is irradiated with light of wavelength $332 n m .$ The retarding potential required to stop the escape of photon electron is:

The correct option is C $2.66 V$
Energy of incident light,
$E (e V) = \frac{12375}{3320} = 3.72 e V$
relation, $E = W_{0} + e V_{0}$
now, $V_{0} = \frac{(E - W_{0})}{e} = \frac{3.72 e V - 1.07 e V}{e}$
$V_{0} = 2.65 V o l t$