Question

The potential diffrerence between the target and the cathode of an X-ray tube is $50$ kV. The current in the tube is $20$ mA. Only 1% of the total energy is emitted as x-rays. (Take hc = $1240$ eV nm)

• A. The maximum frequency of the emitted x-rays is approx $1.2\times {10}^{19}\text{Hz}$.
• B. The heat should be removed at $900W$ to keep the target at constant temperature
• C. A wavelength of 0.2 nm can be emitted by this tube
• D. Electrons are being emitted at the rate of $1.25\times {10}^{17}/\text{sec}$ by the cathode

Answer 1

Answer: (A), (C), (D)

The correct options are
A The maximum frequency of the emitted x-rays is approx $1.2\times {10}^{19}\text{Hz}$.
C A wavelength of 0.2 nm can be emitted by this tube
D Electrons are being emitted at the rate of $1.25\times {10}^{17}/\text{sec}$ by the cathode

Use $E=\frac{hc}{\lambda }$
$\lambda =\frac{hc}{E}$
$\lambda =\frac{1240}{50\times {10}^3}$
Now for frequency
$f=\frac{c}{\lambda }$
where $c$ = velocity of light
$\lambda =$ wavelength
put the values of c and $\lambda$ in formula -
$f=\frac{3\times {10}^8\times 50\times {10}^3}{1240\times {10}^{-9}}$
$=\frac{150}{124}\times {10}^{19}=1.2\times {10}^{19}\text{Hz}.$
$\frac{dQ}{dt}=50\times {10}^3\times 20\times {10}^{-3}=1000W$
$\frac{dQ}{dt}=1000W$
${\lambda }_{\text{min}}=\frac{1240}{50\times {10}^{+3}}=24.8\times {10}^{-3}\text{nm}$
$\frac{dN}{dt}=\frac{20\times {10}^{-3}}{1.6\times {10}^{-19}}=1.25\times {10}^{17}/\text{sec}$