For intensity I of a light of wavelength 5000 A the photoelectron saturation current is 0.40 µA and stopping potential is 1.36 V, the work function of metal is

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2.47 eV

1.36 eV

1.10 eV

0.43 eV

answer is C.

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Detailed Solution

$\begin{array}{l} By using E = W_{o} + K_{\max} \\ \Rightarrow E = \frac{12375}{5000} = 2 .475 eV and K_{\max} = {eV}_{o} = 1 .36 eV \\ So 2 .475 = W_{o} + 1 .36 \Rightarrow W_{o} = 1 .1 eV \end{array}$

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