Question

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

• A. The maximum frequency of the emitted x-rays is approx $1.2\times {10}^{19}Hz$.
• B. The heat should be removed at $900\text{W}$ to keep the target at constant temperature
• C. A wavelength of $0.8\text{nm}$ 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), (D)

The correct options are
A The maximum frequency of the emitted x-rays is approx $1.2\times {10}^{19}Hz$.
D electrons are being emitted at the rate of $1.25\times {10}^{17}\text{/sec}$ by the cathode.

Wave length $\lambda =\frac{1240}{50\times {10}^3}\text{nm}$
Frequency
$f=\frac{c}{\lambda }=\frac{3\times {10}^8\times 50\times {10}^3}{1240\times {10}^{-9}}=\frac{150}{124}\times {10}^{19}=1.2\times {10}^{19}\text{Hz}$
Rate of heat removal
$P=\frac{dQ}{dt}=50\times {10}^3\times 20\times {10}^{-3}=1000\text{W}$
$P_{xrays}=1000\text{W}$
Minimum wavelength
${\lambda }_{min}=\frac{1240}{50\times {10}^{+3}}=24.8\times {10}^{-3}\text{nm}$
Rate of emission of electrons
$\frac{dN}{dt}=\frac{20\times {10}^{-3}}{1.6\times {10}^{-19}}=1.25\times {10}^{17}\text{/sec}$