A metal ball of work function is irradiated by a radiation of wavelength $3100 \overset{\circ}{A} .$ Consider two electrons emitted by the ball moving along X and Y axes having kinetic energies K and $K - 1.5 e V$ respectively, here K is maximum kinetic energy of an emitted photoelectron. If de-Broglie wavelength of electron moving along X-axis is $λ$ , de-Broglie wavelength of electron moving along Y-axis in centre of mass frame of the system is $\frac{4 λ}{\sqrt{x}}$ . Find x.
(Take, $h c = 12400 e V - \overset{o}{A}$ )

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answer is 5.

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Detailed Solution

$A s, K = \frac{12400}{3100} - 2 = 2 e V = K_{1} \Rightarrow K_{2} = K - 1.5 = 0.5$

Since, $λ \propto p \Rightarrow \frac{λ_{1}}{λ_{2}} = \frac{p_{2}}{p_{1}} = \sqrt{\frac{K_{2}}{K_{1}}} = \frac{1}{2} \Rightarrow λ_{2} = 2 λ_{1} = 2 λ$

$v_{C M} = \frac{v_{1} \hat{i} + v_{2} \hat{j}}{2} \Rightarrow v_{2, C M} = v_{2} j - (\frac{v_{1} i + v_{2} j}{2}) = \frac{- v_{1} i + v_{2} j}{2} \Rightarrow λ_{2, C M} = \frac{h}{m \sqrt{\frac{v_{1}^{2} + v_{2}^{2}}{4}}} = \frac{2 λ_{1} λ_{2}}{\sqrt{λ_{1}^{2} + λ_{2}^{2}}} = \frac{2 \cdot λ \cdot 2 λ}{\sqrt{λ^{2} + {(2 λ)}^{2}}} = \frac{4 λ}{\sqrt{5}}$