Question

The potential difference between the, cathode and the target electrode in a coolidge tube is $24.75kV$. 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
$eV=h{v}_{\max }$
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.
$\therefore eV=\frac{hc}{{\lambda }_{\min }}$
$\Rightarrow {\lambda }_{\min }=\frac{hc}{eV}$
Given, $V=24.75kV=24.75\times 10^{3}V$
$h=6.6\times 10^{-34}Js,c=3\times 10^{8}m{s}^{-1},$
$e=1.6\times 10^{-19}C$
$\therefore {\lambda }_{\min }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{1.6\times 10^{-19}\times 24.75\times 10^3}$
$=0.5\times 10^{-10}$
$=0.5\mathring{A}$