.An electron moving with a speed of $5 \times 10^{6} {ms}^{- 1}$ is shot parallel to an electric field of strength $1000 {Vm}^{- 1}$ , arranged so as to retard its motion. The electron travels a distance $s$ in the field before coming momentarily to rest in time $t$ . It is observed that the electric field ends abruptly after $0.8 cm$ and due to this the electron loses a percentage fraction $f$ of its initial energy, then

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$t = 0.03 s$

$f = 11 %$

$s = 0.14 m$

$s = 0.07 m$

answer is B, C, D.

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Detailed Solution

Retardation $a = \frac{force}{mass} = \frac{e E}{m}$

$\Rightarrow a = \frac{1.6 \times 10^{- 19} \times 1000}{9 \times 10^{- 31}} = 1.78 \times 10^{14} {ms}^{- 2}$

Since, $v^{2} - u^{2}$ = 2 a s

$\Rightarrow 0^{2} - {(5 \times 10^{6})}^{2} = 2 \times {(- 1.78 \times 10^{14})}_{S} \Rightarrow s = 0.07 m . N o w, v = v_{0} + a t \Rightarrow 0 = 5 \times 10^{6} - (1.78 \times 10^{14}) t \Rightarrow t = 2.8 \times 10^{- 8} s = 0.03 μ s L o s s o f e n e r g y = w o r k d o n e W = F d = (e L), 1$