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In a photoelectric experiment, threshold wavelength of light of a metal is 380 nm. If wavelength of 280 nm is incidented on metal, the maximum kinetic energy of emitted electron will be: $[E_{(in ev)} = \frac{1237}{λ_{(in nm)}}]$

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15.1 eV

3.0 eV

1.5 eV

4.5 eV

answer is A.

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Detailed Solution

$hv = {hv}_{0} + \frac{1}{2} {mu}^{2} \begin{array}{l} \frac{1}{2} {mu}^{2} = hv - {hv}_{0} = \frac{hc}{λ} - \frac{hc}{λ_{0}} = \frac{1237}{260} - \frac{1237}{380} \\ K . E = \frac{1237 \times 120}{260 \times 380} = 15 .02 eV \end{array}$

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