Question

Crystal diffraction experiments can be performed either by using electrons accelerated through appropriate voltage or by using X-rays. If the wavelength of these probes (electrons or X-rays) is $1\mathring{A}$, estimate which of the two has greater energy?

• A. X-ray
• B. Electron.
• C. both have same energy
• D. cannot be determined by crystal diffraction.

Answer 1

Answer: (A)

For an accelerated electron beam, the de-Broglie matter wave equation states that
$\lambda =\frac{h}{\sqrt{2emV}}=\frac{h}{\sqrt{2mK}}$
$\Rightarrow K=\frac{h^2}{2m{\lambda }^2}$
For X-rays, photon of same wavelength $\lambda =1\mathring{A}$
$E^{\prime }=hv=\frac{hc}{\lambda }$ or
$\frac{K}{E^{\prime }}=\frac{\frac{h^2}{2m{\lambda }^2}}{\frac{hc}{\lambda }}$
or $\frac{K}{E^{\prime }}=\frac{h^2}{2m{\lambda }^2}\times \frac{\lambda }{hc}$
$=\frac{h}{2mc\lambda }$
$\therefore \frac{K}{E^{\prime }}=\frac{h}{2mc\lambda }$
Where $h=6.6\times 10^{-34}Js$,
$m=9\cdot 1\times 10^{-31}kg$
$c=3\times 10^{8}m/s\lambda =1\mathring{A}=10^{-10}m$
$\frac{K}{E^{\prime }}=\frac{6.6\times 10^{-34}}{2\times 9.1\times 10^{-31}\times 3\times 10^8\times 10^{-10}}=\frac{11}{911}$
$\Rightarrow K<E^{\prime }$
$\therefore$ Energy possessed by $X$ -ray is more than electron.