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Q. 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: $\frac{h^{2}}{m_{e} \times e} = 3 \times 10^{- 18} J^{2} s^{2} kg^{- 1} coulomb^{- 1}$

NTA AbhyasNTA Abhyas 2020Structure of Atom

Solution:

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

$V=\frac{1}{2}\times \frac{\left(6.62 \times 1 0^{- 34}\right)^{2}}{9.108 \times 1 0^{- 31} \times \left(1 \times 1 0^{- 10}\right)^{2} \times 1.602 \times 1 0^{- 19}}=150 \, volt$