Question

The wavelength of $K_{\alpha }\text{X-ray}$ of tungsten is $21.3\text{pm}$. It takes $11.3\text{keV}$ to knock out an electron from the $\text{L-shell}$ of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having a tungsten target which allows production of $K_{\alpha }\text{X-ray}$? $(hc=1242\text{eV-nm})$

• A. $53.8\text{kV}$
• B. $74.5\text{kV}$
• C. $89.2\text{kV}$
• D. $69.6\text{kV}$

Answer 1

The correct option is D $69.6\text{kV}$

Given, ${\lambda }_K=21.3\times {10}^{-3}\text{nm}$

Now, $E_K-E_L=\frac{hc}{{\lambda }_K}=\frac{1242\text{eV-nm}}{21.3\times {10}^{-3}\text{nm}}=58.309\text{keV}$

Given, $E_L=11.3\text{keV}$

$\Rightarrow E_K=58.309+11.3=69.609\text{keV}$

Now, minimum accelerating voltage $V$ is calculated by,

$E_K=Ve=69.609\text{keV}$

$\therefore V=69.609\text{kV}$

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