Question

An $\alpha$ -particle is accelerated through a potential difference of V volts from rest.The de-Broglie's was 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 energy due to potential difference $V$ will provide Kinetic energy to $\alpha -$particle.

Mass of $\alpha -$ particle $=4\times 1.67\times {10}^{-27}kg=6.68\times {10}^{-27}kg$

Charge of $\alpha -$ particle $=2\times 1.6\times {10}^{-19}C=3.2\times {10}^{-19}C$

$\frac{1}{2}m{v}^2=qV$

$v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2\times 3.2\times {10}^{-19}V}{6.68\times {10}^{-27}}}=0.978\times {10}^4\sqrt{V}$ $m/s$

de-Broglie wavelength is, $\lambda =\frac{h}{p}=\frac{h}{mv}$

$\lambda =\frac{6.6\times {10}^{-34}}{6.68\times {10}^{-27}\times 0.978\times {10}^4\sqrt{V}}$

$\lambda =\frac{0.101}{\sqrt{V}}$