Q. What is the de-Broglie wavelength of the $\alpha$-particle accelerated through a potential difference $V$
A) $\frac{0.287}{\sqrt{V}}\mathring{\mathrm{A}}$
B) $\frac{12.27}{\sqrt{V}}\mathring{\mathrm{A}}$
C) $\frac{0.101}{\sqrt{V}}\mathring{\mathrm{A}}$
D) $\frac{0.202}{\sqrt{V}}\mathring{\mathrm{A}}$
Explanation
(c) $\lambda = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2m_{\alpha}Q_{\alpha}V}}$
On putting $Q_{\alpha} = 2 \times 1.6 \times 10^{- 19}C$
$m_{\alpha} = 4m_{p} = 4 \times 1.67 \times 10^{- 27}kg \Rightarrow \lambda = \frac{0.101}{\sqrt{V}}\mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.