An electron with kinetic energy $K_{1}$ enters between parallel plates of a capacitor at an angle ' $α$ ' with the plates. It leaves the plates at angle ' $β$ ' with kinetic energy $K_{2}$ . Then the ratio of kinetic energies will be : $\frac{K_{1}}{K_{2}} = \frac{5}{n} \frac{\cos^{2} β}{\cos^{2} α}$ then n is _____

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answer is 5.

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Detailed Solution

From figure, $v_{1} \cos α = v_{2} \cos β$
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$∴ \frac{K_{1}}{K_{2}} = \frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v_{2}^{2}} = \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{\cos^{2} β}{\cos^{2} α}$

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