Q. X-ray of wavelength $\lambda = 2\mathring{\mathrm{A}}$ is emitted from the metal target. The potential difference applied across the cathode and the metal target is
A) 5525 V
B) 320 V
C) 6200 V
D) 3250 V
Explanation
(c) Potential difference $V = \frac{hc}{e\lambda}$
= $\frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 2 \times 10^{- 10}}$
= 6200 V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.