Question

An $X$-ray tube operates on $30kV$. 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
$h{v}_{\max }=eV$ The minimum wavelength corresponding to this maximum energy is
${\lambda }_{\min }=\frac{hc}{eV}$
where $h$ is Planck's constant and $c$ is speed of light.
$\therefore$ ${\lambda }_{\min }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{1.6\times 10^{-19}\times 30\times 10^3}=0.4\mathring{A}$