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An electron with an initial speed of $4.0 \times 10^6 \,ms ^{-1}$ is brought to rest by an electric field. The mass and charge of an electron are $9 \times 10^{-31} \,kg$ and $1.6 \times 10^{-19} \,C$, respectively. Identify the correct statement.
The electron moves from a region of lower potential to higher potential through a potential difference of $11.4 \,\mu V$
The electron moves from a region of higher potential to lower potential through a potential difference of $11.4 \,\mu V$
The electron moves from a region of lower potential to higher potential through a potential difference of $45 \,V$
The electron moves from a region of higher potential to lower potential through a potential difference of $45 \,V$
Solution
(d)
From, $q V=\frac{1}{2} m v^2$
where, $V$ is stopping potential.
$V=\frac{m v^2}{2 q}$
$=\frac{9 \times 10^{-31} \times\left(4 \times 10^6\right)^2}{2 \times 16 \times 10^{-19}}$
$\approx 45 \,V$
So, electron must move across a potential difference of $45 \,V$ from higher to lower potential.
