Q. An important spectral emission line has a wavelength of 21cm. The corresponding photon energy is $(h = 6.62 \times 10^{- 34}$ Js and $c = 3 \times 10^{8}\ ms^{- 1})$
A) $5.9 \times 10^{- 8}$ eV
B) $5.9 \times 10^{- 4}$ eV
C) $5.9 \times 10^{- 6}$ eV
D) $11.8 \times 10^{- 6}$ eV
Explanation
(c) $E = hv/\lambda\ = \frac{hc}{e\lambda}$ (in eV)
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 0.21} = 5.9 \times 10^{- 6}$ eV
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.