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 photoelectrons is

4.81 eV

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3.74 eV

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2.66 eV

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1.07 eV

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Solution

The correct option is C 2.66 eV
Energy of incident light $E (e V) = \frac{12375}{λ} = \frac{12375}{3320} = 3.72 e V$
(332 nm = 3320 Å)
According to the relation $E = W_{0} + e V_{0}$
$\Rightarrow V_{0} = \frac{(E - W_{0})}{e} = \frac{3.72 e V - 1.07 e V}{e} = 2.65 V o l t .$

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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 photoelectrons is

The correct option is C 2.66 eVEnergy of incident light E(eV)=12375λ=123753320=3.72 eV (332 nm = 3320 Å) According to the relation E=W0+eV0 ⇒V0=(E−W0)e=3.72 eV−1.07 eVe=2.65Volt.

The correct option is C 2.66 eV
Energy of incident light $E (e V) = \frac{12375}{λ} = \frac{12375}{3320} = 3.72 e V$
(332 nm = 3320 Å)
According to the relation $E = W_{0} + e V_{0}$
$\Rightarrow V_{0} = \frac{(E - W_{0})}{e} = \frac{3.72 e V - 1.07 e V}{e} = 2.65 V o l t .$