Question

A beam of fast moving electrons having cross-sectional area $A$ falls normally on a flat surface. The electrons are absorbed by the surface and the average pressure exerted by the electrons on this surface is found to be $P$. If the electrons are moving with a speed $v$, then the effective current through any cross-section of the electron beam is

• A. $APe/\left(mv\right)$
• B. $APe/\left(m{v}^2\right)$
• C. $APv/\left(me\right)$
• D. $APm/\left(eV\right)$

Answer 1

Answer: (A)

$APe/\left(mv\right)$

Given,

Cross-sectional area $=A$

Average pressure exerted by the electrons $=P$.

Let the number of electrons in the beam be $n$.

Then
Change in momentum $\Delta mv=n{m}_{e}v$

Forces is given as
$F=PA$
Now, we know

$F\times t=\Delta mv$

$\Rightarrow F\times t=n{m}_{e}v$

$\Rightarrow PA\times t=n{m}_{e}v$

$\Rightarrow t=\frac{n{m}_{e}v}{PA}$

We know,

Current, $i=\frac{q}{t}$

$q=ne$

$\Rightarrow i=\frac{ne}{\left(\frac{n{m}_{e}v}{PA}\right)}$

$\Rightarrow i=\frac{APe}{m_{e}v}$

Hence, the correct answer is OPTION A.