Question

A $\alpha$ - particle is accelerated through a potential difference of V volts from rest. The de-Broglie wavelength associated with it is (in angstrom):

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

Answer 1

Answer: (C)

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

Kinetic Energy gained by the $\alpha$-particle by potential difference $V$ is,
$\frac{p^2}{2m_{\alpha }}=q_{\alpha }V=2eV$,

and $m_{\alpha }=4m_p$

$p^2==16m_{p}eV$

$p=4\sqrt{m_{p}eV}=4\sqrt{(1.67\times {10}^{-27})kg\times (1.6\times 106-19)C\times V}=4\times 1.634\times {10}^{-23}\sqrt{V}$

$\lambda =\frac{h}{p}$, here $h=6.62\times {10}^{-34}J-sec$ is Plank's constant.

$\lambda =\frac{h}{p}=\frac{6.62\times {10}^{-34}}{4\times 1.634\times {10}^{-23}\sqrt{V}}=\frac{0.101}{\sqrt{V}}\stackrel{\circ }{A}$

Option 'C' is correct.