Question

An electron moving with a speed of $5\times {10}^{6}m{s}^{-1}$ is shot parallel to an electric field of strength $1000V{m}^{-1}$, 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 0.85 cm and due to this the electron loses a percentage fraction f of its initial energy, then

• A. s = 0.14 m
• B. s = 0.01 m
• C. $t=0.09\mu s$
• D. $f=11\%$

Answer 1

The correct option is D $f=11\%$

Retardation $a=\frac{force}{mass}=\frac{eE}{m}$
$\Rightarrow a=\frac{1.6\times {10}^{-19}\times 1000}{9\times {10}^{-31}}=1.78\times {10}^{14}m{s}^{-2}$
Since, $v^2-v_0^2=2as$
$\Rightarrow 0^2-(5\times {10}^6{)}^2=2(-1.78\times {10}^{14})s$
$\Rightarrow s=0.07m$
Now, $v=v_0-at$
$\Rightarrow 0=5\times {10}^6-(1.78\times {10}^{14})t$
$\Rightarrow t=2.8\times {10}^{-8}s=0.03\mu s$
Loss of energy = work done W = Fd = (eE)d
$\Rightarrow Fractionf=\frac{eEd}{\frac{1}{2}m{v}^2}=\frac{2eEd}{m{v}^2}$
$\Rightarrow \%Fraction=\frac{2\times {1.6}^{-16}\times 1000\times 0.8\times {10}^{-2}}{9\times {10}^{-31}\times (5\times {10}^6{)}^2}=11\%$