Question

An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to $1.54\mathring{A}$?

• A. $13.6$ volt
• B. $1.54$ volt
• C. $1$ volt
• D. $63.44$ volt

Answer 1

Answer: (D)

$63.44$ volt

For an electron $\frac{1}{2}m{u}^2=eV$; also $\lambda =\frac{h}{mu}$

$\therefore \frac{1}{2}m\frac{h^2}{m^2{\lambda }^2}=eV$

$V=\frac{1}{2}\frac{h^2}{m{\lambda }^{2}e}$

$V=\frac{1\times (6.626\times {10}^{-34}{)}^2}{2\times 9.108\times {10}^{-31}\times (1.54\times {10}^{-10}{)}^2\times 1.602\times {10}^{-19}}$

$=63.44$ volt

Hence, the correct option is $D$