Q. The difference between kinetic energies of photoelectrons emitted from a surface by light of wavelength 2500$\mathring{\mathrm{A}}$ and 5000$\mathring{\mathrm{A}}$ will be
A) 1.61 eV
B) 2.47 eV
C) 3.96 eV
D) $3.96 \times 10^{- 19}eV$
Explanation
(b) $\Delta E = \frac{hc}{\lambda_{1}} - \frac{hc}{\lambda_{2}} = \frac{hc(\lambda_{2} - \lambda_{1})}{\lambda_{1}\lambda_{2}}\ (in\ eV)$
$= \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8} \times (5000 - 2500) \times 10^{- 10}}{2500 \times 5000 \times 10^{- 20} \times 1.6 \times 10^{- 19}}$
$= 2.47\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.