First slide

Photo Electric Effect

Question

In a photoemissive cell with executing wavelength $λ$ , the fastest electron has speed v. If the exciting wavelength is changed to 3 $λ$ /4, the speed of the fastest emitted electron will be

Moderate

A

$v {(3 / 4)}^{1 / 2}$
B

$v {(4 / 3)}^{1 / 2}$
C

$Less than v {(4 / 3)}^{1 / 2}$
D

$Greater than v {(4 / 3)}^{1 / 2}$

Solution

$\begin{array}{l} hv - W_{o} = \frac{1}{2} {mv}_{\max}^{2} \Rightarrow \frac{hc}{λ} - \frac{hc}{λ_{o}} = \frac{1}{2} {mv}_{\max}^{2} \\ \Rightarrow hc (\frac{λ_{o} - λ}{{λλ}_{o}}) = \frac{1}{2} {mv}_{\max}^{2} \Rightarrow v_{\max} \sqrt{\frac{2 hc}{m} (\frac{λ_{o} - λ}{{λλ}_{o}})} \\ When wavelength is λ and velocity is v, then \\ v = \sqrt{\frac{2 hc}{m} (\frac{λ_{o} - λ}{{λλ}_{o}})} . . . . (i) \\ When wavelength is \frac{3 λ}{4} and velocity is v' then \\ v' = \sqrt{\frac{2 hc}{m} [\frac{λ_{o} - (3 λ / 4)}{(3 λ / 4) \times λ_{o}}]} . . (ii) \\ Dividing equation (ii) by (i), we get \\ \frac{v'}{v} = \sqrt{\frac{[λ_{o} (3 λ / 4)]}{\frac{3}{4} {λλ}_{o}} \times \frac{{λλ}_{o}}{λ_{o} - λ}} \\ v' = v {(\frac{4}{3})}^{1 / 2} \sqrt{\frac{[λ_{0} - (3 λ / 4)]}{λ_{o} - λ}} i . e ., v' > {(\frac{4}{3})}^{1 / 2} \end{array}$

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