Question

An $\alpha$-particle is accelerated through a potential difference of V volts from rest. The De-Broglie's wavelength associated with it is:

• A. $\sqrt{\frac{150}{V}}$ A${}^o$
• B. $\frac{0.286}{\sqrt{V}}$ A${}^o$
• C. $\frac{0.101}{\sqrt{V}}$ A${}^o$
• D. $\frac{0.983}{\sqrt{V}}$ A${}^o$

Answer 1

Answer: (C)

$\frac{0.101}{\sqrt{V}}$ A${}^o$

The potential difference is V volts

The kinetic energy is $\frac{1}{2}m{u}^2=eV$

$u=\sqrt{\frac{2eV}{m}}$

The de Broglie wavelength is $\lambda =\frac{h}{mu}=\frac{h}{m\times \sqrt{\frac{2eV}{m}}}$

Substitute values in the above expression, we get

$\lambda =\frac{6.626\times {10}^{-34}}{4\times 1.67\times {10}^{-27}\times \sqrt{\frac{2\times 2\times 1.602\times {10}^{-19}\times V}{4\times 1.67\times {10}^{-27}}}}$

$\lambda =\frac{0.101}{\sqrt{V}}{A}^0$

Hence, the correct option is $C$