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Illuminating the surface of al certain metal alternately with light of wavelengths $λ_{1} = 0.35 μm$ and $λ_{2} = 0.54 μm$ , it was found that the corresponding maximum velocities of photo electrons differ by a factor $η = 2$ . Find the work function of that metal.

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$5.6 eV$

$1.9 eV$

$3.2 eV$

$4.9 eV$

answer is A.

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Detailed Solution

From Einstein's photoelectricity equation, we have

$\frac{1}{2} m v_{1}^{2} = \frac{h c}{λ_{1}} - W$

$\frac{1}{2} m v_{2}^{2} = \frac{h c}{λ_{2}} - W$

Dividing equation (1) with equation (2), and using the given fact that $v_{1} = 2 v_{2}$ , we get

$4 = \frac{\frac{h c}{λ_{1}} - W}{\frac{h c}{λ_{2}} - W}$

$\Rightarrow 3 W = 4 (\frac{h c}{λ_{2}}) - (\frac{h c}{λ_{1}}) = \frac{4 \times 12375}{5400} - \frac{12375}{3500} = 5.63 eV$

$\Rightarrow W = 1.9 eV$

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