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 photo electron is

4.81 e V

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3.74 e V

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2.65 e V

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1.07 e V

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Solution

The correct option is C $2.65 e V$
Energy of incident light
$E (e V) = \frac{12375}{3320} = 3.72 e V$
We know that, $E = W_{0} + {e V}_{0}$
$V_{0} = \frac{(E - W_{0})}{e} = \frac{3.72 e V - 1.07 e V}{e}$
$V_{0} = 2.65$ Volt

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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 photo electron is

The correct option is C 2.65eVEnergy of incident light E(eV)=123753320=3.72eVWe know that, E=W0+eV0V0=(E−W0)e=3.72eV−1.07eVeV0=2.65 Volt

The correct option is C $2.65 e V$
Energy of incident light
$E (e V) = \frac{12375}{3320} = 3.72 e V$
We know that, $E = W_{0} + {e V}_{0}$
$V_{0} = \frac{(E - W_{0})}{e} = \frac{3.72 e V - 1.07 e V}{e}$
$V_{0} = 2.65$ Volt