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Given that a photon of light of wavelength 10,000 $Å$ has an energy equal to 1.23 eV. When light of wavelength 5000 $Å$ and intensity I_o falls on a photoelectric cell,
the saturation current is 0.40 x 10^-6 A and the stopping potential is 1.36 V; then work function is

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equal to 1.1 eV

equal to 1.46 eV

equal to (1.56/2) eV

equal to 1.36 eV

answer is C.

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Detailed Solution

$E = \frac{12400}{λ}$

$\frac{E_{2}}{E_{1}} = \frac{λ_{1}}{λ_{2}} \Rightarrow \frac{E_{2}}{1.23} = \frac{5000}{10000}$

$E_{2} = 2.46 e V$

$E = K E + W$

$W = E - e V_{0} = 2.46 - 1.36 = 1.1 e V$

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