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 $1.0 Å$ . The value of 'V' is
Given: $\frac{h^{2}}{m_{e} \times e} = 3 \times 10^{- 18} J^{2} s^{2} {kg}^{- 1} {coulomb}^{- 1}$

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We know that 1 2 m u 2 = eV and λ = h mu or u = h m λ or u 2 = h 2 m 2 λ 2 1 2 m × h 2 m 2 λ 2 = eV or V = 1 2 m × h 2 m 2 λ 2 × e = 1 2 × h 2 m λ 2 × e Substituting the values, we get V = 1 2 × 6.62 × 1 0 - 34 2 9.108 × 1 0 - 31 × 1 × 1 0 - 10 2 × 1.602 × 1 0 - 19 = 150 v o l t

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