Q. The de-Broglie wavelength relates to applied voltage ror $\alpha$-particles as
A) $\lambda = \frac{12.3A^{{^\circ}}}{\sqrt{V}}$
B) $\lambda = \frac{0.286}{\sqrt{V}}A^{{^\circ}}$
C) $\lambda = \frac{0.101}{\sqrt{V}}A^{{^\circ}}$
D) $\lambda = \frac{0.856}{\sqrt{V}}A^{{^\circ}}$
Explanation
(c)
The de-Broglie wavelength associated with the charged particle as
For electron, $\lambda = \frac{12.27}{\sqrt{V}}\mathring{\mathrm{A}}$
For proton, $\lambda = \frac{0.286}{\sqrt{V}}\mathring{\mathrm{A}}$
For $\alpha$-particles, $\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.