Question

The potential difference between the, cathode and the target electrode in a coolidge tube is $24.75 \,kV$. The minimum wavelength of the emitted X-rays is

• A. $ 0.1 \,\mathring{A}$
• B. $ 0.5 \,\mathring{A}$
• C. $ 1 \,\mathring{A}$
• D. $ 5 \,\mathring{A}$

Answer 1

Answer: (B)

When electrons accelerated through a potential difference $V$ strike a target, the maximum frequency of the emitted $X$-rays is given by
$e V=h v_{\max }$
where, $e$ is charge of the electron and $h$ the Planck's constant.
But $v_{\max }=\frac{c}{\lambda_{\min }}$, where $c$ is the speed of light and $\lambda_{\min }$ the minimum wavelength.
$\therefore e V =\frac{h c}{\lambda_{\min }} $
$\Rightarrow \lambda_{\min } =\frac{h c}{e V}$
Given, $ V =24.75\, kV =24.75 \times 10^{3} V $
$h =6.6 \times 10^{-34} Js , c=3 \times 10^{8} ms ^{-1},$
$ e =1.6 \times 10^{-19} C $
$ \therefore \lambda_{\min } =\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 24.75 \times 10^{3}}$
$=0.5 \times 10^{-10} $
$=0.5\,\mathring{A}$