Energy levels \(A\), \(B\), \(C\) of a certain atom correspond to increasing values of energy \((E_A<E_B<E_C)\). If \(\lambda_1\), \(\lambda_2\) and \(\lambda_3\) are the wavelengths of radiations corresponding to the transitions \(C\to B\), \(B\to A\) and \(C\to A\) respectively, the relation between these wavelengths can be written as
Show Hint
For atomic transitions:
\[
\Delta E=\frac{hc}{\lambda}
\]
Whenever energy differences add,
\[
\frac{1}{\lambda}
\]
values add, not the wavelengths themselves.
This is analogous to the Rydberg formula used in atomic spectra.
Concept:
For an electronic transition,
\[
\Delta E=\frac{hc}{\lambda}
\]
where \(h\) is Planck's constant and \(c\) is the speed of light.
Also,
\[
E_{CA}=E_{CB}+E_{BA}
\]
because the total energy difference between levels \(C\) and \(A\) equals the sum of the intermediate energy differences.
Step 1:Write the energy relations.
For transition \(C\to B\),
\[
E_C-E_B=\frac{hc}{\lambda_1}
\]
For transition \(B\to A\),
\[
E_B-E_A=\frac{hc}{\lambda_2}
\]
For transition \(C\to A\),
\[
E_C-E_A=\frac{hc}{\lambda_3}
\]
Step 2: Use energy conservation.
\[
(E_C-E_A)
=
(E_C-E_B)+(E_B-E_A)
\]
Substituting,
\[
\frac{hc}{\lambda_3}
=
\frac{hc}{\lambda_1}
+
\frac{hc}{\lambda_2}
\]
Cancelling \(hc\),
\[
\frac{1}{\lambda_3}
=
\frac{1}{\lambda_1}
+
\frac{1}{\lambda_2}
\]
Step 3: Simplify.
\[
\frac{1}{\lambda_3}
=
\frac{\lambda_1+\lambda_2}
{\lambda_1\lambda_2}
\]
Hence,
\[
{
\lambda_3=
\frac{\lambda_1\lambda_2}
{\lambda_1+\lambda_2}
}
\]
Step 4: State the answer.
\[
{
\lambda_3=
\frac{\lambda_1\lambda_2}
{\lambda_1+\lambda_2}
}
\]
Hence, the correct option is
\[
{(B)}
\]
