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Q.

The photoelectric threshold wavelength for silver is $λ_{0} .$ The energy of the electron rejected from the surface of silver by an incident wavelength $λ (λ < λ_{0})$ will be:

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a

$\frac{h}{c} (\frac{λ_{0} - λ}{λ λ_{0}})$

b

$\frac{hc}{λ_{0} - λ}$

c

$hc (\frac{λ_{0} - λ}{λ λ_{0}})$

d

$hc (λ_{0} - λ)$

answer is D.

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Detailed Solution

$KE = hc [\frac{1}{λ} - \frac{1}{λ_{0}}]$

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