During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is \(2000 Å\) and it becomes \(6000 Å\) 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 _______.

\(\frac{h c}{\lambda_1}=E_0 z^2\left(\frac{1}{n_c^2}-\frac{1}{n_A^2}\right)\) ....(i)
And \(\frac{h c}{\lambda_2}=E_0 z^2\left(\frac{1}{n_c^2}-\frac{1}{n_B^2}\right)\)
So for \(A\) and \(B\) ....(ii)
\(\frac{h c}{\lambda_3}=E_0 z^2\left(\frac{1}{n_B^2}-\frac{1}{n_A^2}\right)\)
Clearly subtracting equation (ii) from equation (i)
\(\begin{aligned}
& h c\left[\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right]=E_0 z^2\left[\frac{1}{n_B^2}-\frac{1}{n_A^2}\right]=\frac{h c}{\lambda_3} \\
& \Rightarrow \frac{1}{\lambda_3}=\frac{1}{\lambda_1}-\frac{1}{\lambda_2} \Rightarrow \frac{1}{\lambda_3}=\frac{(6000-2000)}{6000 \times 2000}=\frac{1}{3000} \\
& \lambda_3=3000 Å
\end{aligned}\)