Q. If the energy of a photon corresponding to a wavelength of $6000\ \mathring{\mathrm{A}}$ is $3.32 \times 10^{- 19}J,$ the photon energy for a wavelength of $4000\ \mathring{\mathrm{A}}$ will be
A) $1.4\ eV$
B) $4.9\ eV$
C) $3.1\ eV$
D) $1.6\ eV$
Explanation
(c) $E = \frac{hc}{\lambda} \Rightarrow \frac{E_{1}}{E_{2}} = \frac{\lambda_{2}}{\lambda_{1}} \Rightarrow \frac{3.32 \times 10^{- 19}}{E_{2}} = \frac{4000}{6000}$
$\Rightarrow E_{2} = 4.98 \times 10^{- 19}J = 3.1\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.