Light of wavelength $λ$ and photon energy 3 eV falling on a metal surface produces photoelectrons of maximum velocity v. If the wavelength of the incident light is changed to $\frac{3}{4} λ$ and the maximum velocity of electron is doubled, the work function of the metal is nearly $\frac{x}{3} e V$ . Then the value of $x$ is ………….

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answer is 8.

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Detailed Solution

$\begin{array}{l} h ν = \frac{1}{2} m v^{2} + W \Rightarrow \frac{1}{2} m v^{2} = h ν - W \\ K E_{1} = E_{1} - W \dots \dots \dots . . . (1); \\ K E_{2} = E_{2} - W \dots \dots \dots . . . . (2) \\ \frac{(1)}{(2)} = \frac{K E_{1}}{K E_{2}} = \frac{E_{1} - W}{E_{2} - W} \\ \Rightarrow \frac{\frac{1}{2} m v^{2}}{\frac{1}{2} m (2 v)^{2}} = \frac{E_{1} - W}{E_{2} - W} \\ \Rightarrow \frac{1}{4} = \frac{E_{1} - W}{E_{2} - W} \\ \Rightarrow E_{2} - W = 4 E_{1} - 4 W \\ \Rightarrow 4 E_{1} - E_{2} = 3 W [W = \frac{4 E_{1} - E_{2}}{3}] \\ \Rightarrow E_{1} λ_{1} = E_{2} λ_{2} [E = \frac{h c}{λ}] \\ \Rightarrow E_{2} = \frac{E_{1} λ_{1}}{λ_{2}} = 3 \times \frac{λ}{\frac{3}{4} λ} = 4 eV \\ \Rightarrow W = \frac{4 E_{1} - E_{2}}{3} = \frac{4 \times 3 - 4}{3} = \frac{8}{3} eV \end{array}$

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