During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is $2000 \mathop A\limits^o$ and it becomes $6000 \mathop A\limits^o$ 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

<p>$$ \frac{1}{\lambda_{AC}} = \frac{E_A - E_C}{hc}, \quad \frac{1}{\lambda_{BC}} = \frac{E_B - E_C}{hc} $$</p> <p>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</p> <p>$E_A - E_B = (E_A - E_C) - (E_B - E_C)$</p> <p>Expressing these energy differences in terms of wavelengths,</p> <p>$\frac{hc}{\lambda_{AB}} = \frac{hc}{\lambda_{AC}} - \frac{hc}{\lambda_{BC}}$</p> <p>Cancelling out $hc$, we obtain</p> <p>$\frac{1}{\lambda_{AB}} = \frac{1}{\lambda_{AC}} - \frac{1}{\lambda_{BC}}$</p> <p>Substitute the given values:</p> <p>$\frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000}$</p> <p>Finding a common denominator:</p> <p>$\frac{1}{\lambda_{AB}} = \frac{3 - 1}{6000} = \frac{2}{6000} = \frac{1}{3000}$</p> <p>Thus, the wavelength for the A to B transition is</p> <p>$\lambda_{AB} = 3000 \, \mathring{A}.$</p> <p>This matches Option D.</p>