Q. The atomic transition gives rise to the radiation of frequency $10^{4}\ MHz$. The change in energy per mole of atoms taking place would be
A) $6.62 \times 10^{- 30}J$
B) $5.32 \times 10^{- 28}J$
C) $6.62 \times 10^{- 20}J$
D) 3.99 $J$
Explanation
(d)
$E = Nhv$
$= 6.023 \times 10^{23} \times 6.626 \times 10^{- 34} \times 10^{4} \times 10^{6}$
$= 3.99$ J
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.