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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