Question

An $X$-ray tube operates on $30\, kV$. The minimum wavelength emitted is : $ (h=6.6\times {{10}^{-19}}\,j-s,c=3\times {{10}^{8}}m/s, $ $ e=1.6\times {{10}^{-19}}C) $

• A. $ 6.6\,\mathring{A}$
• B. $ 1.2\,\mathring{A} $
• C. $ 0.133\,\mathring{A}$
• D. $ 0.4\,\mathring{A}$

Answer 1

Answer: (D)

For an accelerating voltage $V$, the maximum $X$-ray photon energy is given by
$h v_{\max }=e V$ The minimum wavelength corresponding to this maximum energy is
$\lambda_{\min }=\frac{h c}{e V}$
where $h$ is Planck's constant and $c$ is speed of light.
$\therefore $ $\lambda_{\min }=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 30 \times 10^{3}}=0.4 \,\mathring{A}$