Q. During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is $2000\,\text{\AA}$ and it becomes $6000\,\text{\AA}$ when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is
A) $4000\,\text{\AA}$
B) $2000\,\text{\AA}$
C) $3000\,\text{\AA}$
D) $6000\,\text{\AA}$
Explanation
[Solution by Dr. Abubacker] $E_{AB}=E_{AC}-E_{BC}$, so $\frac1{\lambda_{AB}}=\frac1{2000}-\frac1{6000}=\frac2{6000}$, giving $\lambda_{AB}=3000\,\text{\AA}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.