Q. The photoelectric threshold of Tungsten is 2300$\mathring{\mathrm{A}}$. The energy of the electrons ejected from the surface by ultraviolet light of wavelength 1800$\ \mathring{\mathrm{A}}$ is $(h = 6.6 \times 10^{- 34}$ J-s)
A) 0.15 eV
B) 1.5 eV
C) 15 eV
D) 150 eV
Explanation
(b) $E_{k} = \frac{hc}{e}\left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)(in\ eV)$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}}\ \left( \frac{10^{10}}{1800} - \frac{10^{10}}{2300} \right) = 1.5\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.