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
$\lambda=\frac{h}{m v}$
$\frac{1}{2} mv ^{2}= K . E$
$\frac{1}{2} m v^2 \times 2 m=2 m(K . E)=m^2 v^2$
$\sqrt{2 m \cdot( K \cdot E )}= mv$
$\lambda=\frac{ h }{\sqrt{2 m ( K \cdot E )}}$
$=\frac{ h }{\sqrt{2 m _{\alpha} \cdot 2 eV }}$
$=\frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.66 \times 10^{-27} \times 4 \times 1.6 \times 10^{-19} \times 2 \times V }}$
$=\frac{6.626 \times 10^{-34}}{\sqrt{ V } \times 10^{-23} \times 6.519}$
$=\frac{1.018 \times 10^{-11}}{\sqrt{ V }}$
$=\frac{0.1018}{\sqrt{ V }}\mathring{A}$