Question

An electromagnetic wave of wavelength ‘λ’ is incident on a photosensitive surface of negligible work function. If ‘m’ mass is of photoelectron emitted from the surface that has de-Broglie wavelength λd, then

• A. $\lambda=(\frac{2h}{mc})\lambda d^2$
• B. $\lambda=(\frac{2m}{hc})\lambda d^2$
• C. $\lambda_d=(\frac{2mc}{h})\lambda^2$
• D. $\lambda=(\frac{2mc}{h})\lambda d^2$

Answer 1

Answer: (D)

$\lambda=(\frac{2mc}{h})\lambda d^2$

Answer 2

The correct answer is option (D): $\lambda=(\frac{2mc}{h})\lambda d^2$