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Q.

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 $K_{1} : K_{2}$ will be:

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a

$\frac{\sin^{2} β}{\cos^{2} α}$

b

$\frac{\cos β}{\cos α}$

c

$\frac{\cos^{2} β}{\cos^{2} α}$

d

$\frac{\cos β}{\sin α}$

answer is B.

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Detailed Solution

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$\begin{array}{l} C o m p o n e n t s o f v e l o c i t y p e r p e n d i c u l a r t o e l e c t r i c f i e l d w i l l r e m a i n s a m e . \\ \Rightarrow u_{1} \cos α = u_{2} \cos β \\ \Rightarrow \frac{u_{1}}{u_{2}} = \frac{\cos β}{\cos α} \\ R a t i o = \frac{K_{1}}{K_{2}} = \frac{\frac{1}{2} m u_{1}^{2}}{\frac{1}{2} m u_{2}^{2}} = {(\frac{u_{1}}{u_{2}})}^{2} = \frac{\cos^{2} β}{\cos^{2} α} \end{array}$

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An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle ‘α’ with the plates. It leaves the plates at angle ‘β’ with kinetic energy K2. Then the ratio of kinetic energies K1:K2 will be: