What distance must an electron move in a uniform potential gradient $200\;V/cm$ in order to gain kinetic energy $3.2 * 10 ^{-18}J$?
Given,
Potential Gradient $(E)$ = $200 V/cm$ = $200 * 10^3 \;V/m$
Kinetic Energy $(E_k)$
Charge of Electron $(q)$ = $1.6 * 10^{-19}\; C$
Mass of Electron $(m)$ = $9.1 * 10 ^{-31}\;kg$
Distance $(S)$ = ?
We know that,
$F = m.a$
⇒ $a$ = $ \frac{F}{m}$ = $\frac{q.E}{m}$ = $\frac{1.6 * 10^{-19}\; * \; 200 * 10^3}{9.1 * 10 ^{-31}}$ = $ 3.5 * 10^{15}\; m/s^2$
Now,
$E_K = $$\frac{1}{2}$$m.v^2$
⇒ $v^2 = $$\frac{2. E_k}{m}$ = $\frac{2 \; * \; (3.2 * 10^{-18})}{(9.1 * 10^{-31})}$ = $7 * 10^{12}\;m/s$
Again, to Calculate the distance,
$v^2 = u^2 + 2as$ [∵ $u = 0$]
$S = $$\frac{v^2}{2a}$ = $\frac{7 * 10^{12}}{2 * (3.5 * 10^{15})}$ = $0.001\;m$
Hence,
The required distance is $0.001\; m$.
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