Question

An X-ray tube operates at $40\text{kV}$. Suppose the electron converts $70\%$ of its energy into a photon at each collision. Find the lowest three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.$(hc=1242\text{eV-nm})$

• A. $44\text{pm},148\text{pm},493\text{pm}$
• B. $148\text{pm},493\text{pm},1399\text{pm}$
• C. $20\text{pm},60\text{pm},499\text{pm}$
• D. $20\text{pm},80\text{pm},499\text{pm}$

Answer 1

The correct option is A $44\text{pm},148\text{pm},493\text{pm}$

Given, applied voltage, $V=40\text{kV}=40\times {10}^3\text{V}$

$\therefore \text{Energy Supplied}\left(E\right)=eV=40\times {10}^3\text{eV}$

Now, $\text{Energy Utilized}\left(E_U\right)=\frac{70}{100}\times 40\times {10}^3=28\times {10}^3\text{eV}$

$\Rightarrow \lambda =\frac{hc}{E}=\frac{1242\text{eV-nm}}{28\times {10}^3\text{eV}}=44.35\times {10}^{-3}\text{nm}=44.35\text{pm}$.

For second wavelength,

$E_2=70\%\text{of left over energy}=\frac{70}{100}(E-E_U)$

$=\frac{70}{100}\times (40-28){10}^3=84\times {10}^2\text{eV}$.

${\lambda }_2=\frac{hc}{E}=\frac{1242\text{eV-nm}}{28\times {10}^3\text{eV}}$

$=147.86\times {10}^{-3}\text{nm}=147.86\text{pm}\approx 148\text{pm}$.

For third wavelength,

$E_3=\frac{70}{100}(E-E_U-E_2)=\frac{70}{100}(40-28-8.4)\times {10}^3=25.2\times {10}^2\text{eV}$

${\lambda }_3=\frac{hc}{E}=\frac{1242\text{eV-nm}}{25.2\times {10}^2}$

$=49.2857\times {10}^{-2}\text{nm}\approx 493\text{pm}$

Hence, $\left(A\right)$ is the correct answer.

Why this question? This question gives an idea to calculate different wavelength produced by the X-ray tube.