Question

The de-Broglie wavelength of an $\alpha$ -particles at a voltage $V$ 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. $\lambda =\frac{0.856}{\sqrt{\nabla }}\mathring{A}$

Answer 1

Answer: (C)

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