QJEEM25Ph9131: JEE Main 2025 Physics - Class 12 JEEM25Ph9131 Solution: During the transition of electron from state A to state C of a…

$\frac{1}{\lambda_{AC}} = \frac{E_A - E_C}{hc}, \quad \frac{1}{\lambda_{BC}} = \frac{E_B - E_C}{hc}$

Since the electron transitions from state A to state C in two steps (A to B and then B to C), the energy difference for the transition from A to B is given by

$E_A - E_B = (E_A - E_C) - (E_B - E_C)$

Expressing these energy differences in terms of wavelengths,

$\frac{hc}{\lambda_{AB}} = \frac{hc}{\lambda_{AC}} - \frac{hc}{\lambda_{BC}}$

Cancelling out $hc$ , we obtain

$\frac{1}{\lambda_{AB}} = \frac{1}{\lambda_{AC}} - \frac{1}{\lambda_{BC}}$

Substitute the given values:

$\frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000}$

Finding a common denominator:

$\frac{1}{\lambda_{AB}} = \frac{3 - 1}{6000} = \frac{2}{6000} = \frac{1}{3000}$

Thus, the wavelength for the A to B transition is

$\lambda_{AB} = 3000 \, \mathring{A}.$

This matches Option D.