Dual Nature — Easy JEE physics MCQ
The energy of a photon of wavelength $620$ nm is (Take $hc = 1240$ eV·nm):
- A. $2$ eV
- B. $1$ eV
- C. $3$ eV
- D. $0.5$ eV
Solution
$E = \frac{hc}{\lambda} = \frac{1240}{620} = 2$ eV
PHYSICS
easy • PYQ Reworded
Question
Read carefully, then pick the best option.
The energy of a photon of wavelength nm is (Take eV·nm):
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