Question

The potential difference applied to an X-ray tube is $5$ kV and the current through it is $3.2$ mA. What is the number of electrons striking the target per second ?

Answer 1

solution We know that,

$I=\frac{△\theta }{△t}$

Here, $△\theta =$ (no of electrons striking in

time interval $△t$) (electronic charge)

$I=3.2mA,△t=Is$ and electronic

charge = $1.0\times {10}^{-19}c$ ,so,

$3.2\times {10}^{-3}=\frac{△n}{(1.6\times {10}^{-19}{)}^{-1}}\left(1\right)$

$\frac{3.2\times {10}^{-3}}{1.6\times {10}^{19}}=△n$

$△n=2\times {10}^{16}$

where $△n\to$ No. of electrons striking

per sec.