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Q.

If $λ_{\circ}$ & be $λ$ threshold wavelength & wavelength of incident light, the velocity of photo electron ejected from the metal Surface is

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a

$\sqrt{\frac{2 hc}{m} (λ_{\circ} - λ)}$

b

$\sqrt{\frac{2 h}{m} (\frac{1}{λ_{\circ}} - \frac{1}{λ})}$

c

$\sqrt{\frac{2 h}{m} (λ_{\circ} - λ)}$

d

$\sqrt{\frac{2 hc}{m} (\frac{λ_{\circ} - λ}{{λλ}_{\circ}})}$

answer is C.

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Detailed Solution

48. K.E of the ejected electron is given by equation $hυ = {hυ}_{0} + \frac{1}{2} {mv}^{2}$
$\frac{1}{2} {mv}^{2} = \frac{hc}{λ} - \frac{hc}{λ_{0}} = hc (\frac{λ_{0} - λ}{{λλ}_{0}})$ $v = \sqrt{\frac{2 hc}{m} (\frac{λ_{0} - λ}{λ_{0} λ})}$

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