Q. If the energy difference between the ground state of an atom and its excited state is $4.4 \times 10^{- 4}J$, the wavelength of photon required to produce the transition
A) $2.26 \times 10^{- 12}m$
B) $1.13 \times 10^{- 12}m$
C) $4.52 \times 10^{- 16}m$
D) $4.52 \times 10^{- 12}m$
Explanation
(d)
$\mathrm{\Delta}E = hv = \frac{hc}{\lambda}$
$\lambda = \frac{hc}{\mathrm{\Delta}E} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{4.4 \times 10^{- 14}}$
$= 4.52 \times 10^{- 12}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.