Question

The electric current in an $\text{X}$- ray tube (from the target to the filament) operating at $40\text{kV}$ is $10\text{mA}$. Assume that on an average, $1\%$ of the total kinetic energy of the electrons hitting the target are converted into $X-\text{rays}$. The total power $\left(\text{in W}\right)$ emitted as $\text{X}$-rays is (Integer only)

Answer 1

Given:
$V=40\text{kV};i=10\text{mA}$

Let $E$ be the total kinetic energy per second, so the energy per second for $\text{X}$-rays,

$E_X=1\%\text{of}E$

$i=\frac{Q}{t}=\frac{Ne}{t}=ne$

$\Rightarrow n=\frac{{10}^{-2}}{1.6\times {10}^{-19}}=0.625\times {10}^{17}\text{electrons/second}$

$KE$ of one electron,

$E_e=eV=1.6\times {10}^{-19}\times 40\times {10}^3=6.4\times {10}^{-15}\text{J}$

So, total kinetic energy per second,

$E=n{E}_e=0.625\times {10}^{17}\times 6.4\times {10}^{-15}$

$\therefore E=4\times {10}^2\text{J}$

$\Rightarrow E_X=\frac{1}{100}\times 400=4\text{J in one second}$

$\therefore$ Power emitted as $\text{X}$- ray,

$E_X=4\text{W}$

Correct answer : $4$