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{2 e m V}}=\frac{h}{\sqrt{2 m K}} $
$\Rightarrow K=\frac{h^{2}}{2 m \lambda^{2}}$
For X-rays, photon of same wavelength $\lambda=1 \,\mathring{A}$
$E'=h v=\frac{h c}{\lambda}$ or
$ \frac{K}{E'}=\frac{\frac{h^{2}}{2 m \lambda^{2}}}{\frac{h c}{\lambda}}$
or $\frac{K}{E'}=\frac{h^{2}}{2 m \lambda^{2}} \times \frac{\lambda}{h c}$
$=\frac{h}{2 m c \lambda}$
$\therefore \frac{K}{E'}=\frac{h}{2 m c \lambda}$
Where $h=6.6 \times 10^{-34} J s $,
$ m=9 \cdot 1 \times 10^{-31} kg$
$c=3 \times 10^{8} m / s \lambda=1 \,\mathring{A}=10^{-10} m$
$\frac{K}{E'}=\frac{6.6 \times 10^{-34}}{2 \times 9.1 \times 10^{-31} \times 3 \times 10^{8} \times 10^{-10}}=\frac{11}{911}$
$\Rightarrow K < E'$
$\therefore $ Energy possessed by $X$ -ray is more than electron.