Q. An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) : ($m_e = 9 \times 10^{-31}$ kg)
A) $12.2 \times 10^{-13}$ m
B) $12.2 \times 10^{-12}$ m
C) $12.2 \times 10^{-14}$ m
D) 12.2 nm
Explanation
[Solution] $\lambda = \frac{12.27}{\sqrt{V}}$ Å $= 0.1227$ Å $= 12.2 \times 10^{-12}$ m. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.