Question

An electron beam can undergo diffraction by crystals. The potential of 'V' volt should a beam of electrons be accelerated so that its wavelength becomes equal to $1.0\mathring{A}$ . The value of 'V' is

Given: $\frac{h^2}{m_e\times e}=3\times 10^{-18}{J}^2{s}^{2}k{g}^{-1}coulom{b}^{-1}$

Answer 1

Answer: 150

We know that

$\frac{1}{2}\text{m}{\text{u}}^2=\text{eV}$

and $\lambda =\frac{h}{\text{mu}}$ or $u=\frac{h}{m\lambda }$ or $u^2=\frac{h^2}{m^2{\lambda }^2}$

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

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

Substituting the values, we get

$V=\frac{1}{2}\times \frac{{\left(6.62\times 10^{-34}\right)}^2}{9.108\times 10^{-31}\times {\left(1\times 10^{-10}\right)}^2\times 1.602\times 10^{-19}}=150volt$