Q 12-12-035JEE MainJEE Main 2025 (24 Jan, Shift 1)Easy
During the transition of an electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is $2000\ \text{Å}$ and it becomes $6000\ \text{Å}$ 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
Answer: (C) $3000\ \text{Å}$
Energies add: $E_{AC} = E_{AB} + E_{BC}$, and $E = hc/\lambda$, so
$$\frac{1}{\lambda_{AC}} = \frac{1}{\lambda_{AB}} + \frac{1}{\lambda_{BC}}$$
$$\frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000} = \frac{2}{6000} \Rightarrow \lambda_{AB} = 3000\ \text{Å}$$
Solution by Sreeraj P, M.Sc Physics