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Q.

Radiation of wavelength $λ$ is incident on a photocell. The fastest emitted electron has speed $v$ . If the wavelength of the incident radiation is changed to $\frac{3 λ}{4}$ , the speed of the fastest emitted electron will be,

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a

$= v {(\frac{3}{5})}^{\frac{1}{2}}$

b

$> v {(\frac{4}{3})}^{\frac{1}{2}}$

c

$< v {(\frac{4}{3})}^{\frac{1}{2}}$

d

$= v {(\frac{4}{3})}^{\frac{1}{2}}$

answer is B.

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Detailed Solution

$Initially, \frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ_{0} When wavelength changes to 3λ/4, \frac{1}{2} m v'^{2} = \frac{4 h c}{3 λ} - ϕ_{0} \Rightarrow \frac{1}{2} m v'^{2} = \frac{4 h c}{3 λ} - \frac{4 ϕ_{0}}{3} + \frac{ϕ_{0}}{3} \Rightarrow \frac{1}{2} m v'^{2} = \frac{4}{3} (\frac{h c}{λ} - ϕ_{0}) + \frac{ϕ_{0}}{3}$

$\Rightarrow \frac{1}{2} m v'^{2} = \frac{4}{3} (\frac{1}{2} m v^{2}) + \frac{ϕ_{0}}{3}$

Since $\frac{ϕ_{0}}{3} > 0$ ,

$\frac{1}{2} m v'^{2} > \frac{4}{3} (\frac{1}{2} m v^{2})$

$\Rightarrow v' > v {(\frac{4}{3})}^{\frac{1}{2}}$

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