Question

The de-Broglie wavelength of an $\alpha $ -particles at a voltage $\text{V}$ is
(Given that $\alpha $ -particle has $2$ units positive charge and $4$ units mass)

• A. $\lambda =\frac{12 . 3}{\sqrt{V}}\mathring{A} $
• B. $\lambda = \frac{0 . 286}{\sqrt{V}} \mathring{A} $
• C. $\lambda =\frac{0 . 101}{\sqrt{V}}\mathring{A} $
• D. $\lambda = \frac{0 . 856}{\sqrt{V}} \mathring{A} $

Answer 1

Answer: (C)

The de-Broglie wavelength associated with the charged particle as
$\lambda=\frac{ h }{ mv }$
$\frac{1}{2} mv ^{2}= K \cdot E$
$\frac{1}{2} mv ^{2} \times 2 m =2 m ( K \cdot E )= m ^{2} v ^{2}$
$\sqrt{2 m \cdot( K \cdot E )}= mv$
$\lambda=\frac{ h }{\sqrt{2 m ( K . E )}}$
$=\frac{h}{\sqrt{2 m _{\alpha} \cdot 2 eV }}$
$=\frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.66 \times 10^{-27} \times 4 \times 1.6 \times 10^{-19} \times 2 \times V}}$
$=\frac{6.626 \times 10^{-34}}{\sqrt{V} \times 10^{-23} \times 6.519}$
$=\frac{1.018 \times 10^{-11}}{\sqrt{V}}$
$=\frac{0.1018}{\sqrt{V}} \AA$