Question 102 of 239

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

A$4000 Å$B$2000 Å$C$3000 Å$D$6000 Å$

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Answer: (C) $3000 Å$

$\frac{h c}{\lambda_1}=E_0 z^2\left(\frac{1}{n_c^2}-\frac{1}{n_A^2}\right)$ ....(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)
$\begin{aligned}
& h c\left[\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right]=E_0 z^2\left[\frac{1}{n_B^2}-\frac{1}{n_A^2}\right]=\frac{h c}{\lambda_3} \\
& \Rightarrow \frac{1}{\lambda_3}=\frac{1}{\lambda_1}-\frac{1}{\lambda_2} \Rightarrow \frac{1}{\lambda_3}=\frac{(6000-2000)}{6000 \times 2000}=\frac{1}{3000} \\
& \lambda_3=3000 Å
\end{aligned}$