Question

(A) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be $1.76\times {10}^{11}C$ $k{g}^{-1}$

(B) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?

Answer 1

Work done by the electric fied = $Ve$

Applying work - Energy theorem,

$W_{EF}=△KE$

$\Rightarrow \frac{m{v}^2}{2}-m\frac{(o{)}^2}{2}=Ve$

$\Rightarrow V=\sqrt{2V\left(\frac{e}{m}\right)}=\sqrt{500\times 2\times 1.76\times {10}^{11}}$

$v=1.326\times {10}^{7}m/s$

For $V=10MV={10}^{7}V$

$V=\sqrt{2v\left(\frac{e}{m}\right)}=\sqrt{2\times {10}^7\times 1.76\times {10}^{11}}$

$v=1.87\times {10}^{9}m/s$

Which can't be possible as the velocity of electron is greater than the $\left(vs\right)$ speed of light $(3\times {10}^{8}m/s)$

Lets find $V$ corresponding $V=Vs$

So, $3\times {10}^8=\sqrt{2V(1.7b\times {10}^{11})}$

$\Rightarrow =\frac{(3\times {10}^8{)}^2}{2\times 1.7b\times {10}^{11}}=255.68KV$

So, new formula $\Rightarrow ve\sqrt{V\left(\frac{e}{m}\right)},V\le 255.68kV$