Question

Due to transitions among its first three energy levels, hydrogenic atom emits radiation at three discrete wavelengths $\lambda_{1}\lambda _{2}$ and $\lambda_{3}\left(\lambda_{1} <\lambda_{2}< \lambda_{3} \right)$. Then

• A. $\lambda_{1}=\lambda_{2}+\lambda_{3}$
• B. $\lambda_{1}+\lambda =\lambda_{3}$
• C. $\frac{1}{\lambda_{1} }+\frac{1}{\lambda_{2} }=\frac{1}{\lambda_{3} }$
• D. $\frac{1}{\lambda_{1} }=\frac{1}{\lambda_{2} }+\frac{1}{\lambda_{3} } $

Answer 1

Answer: (D)

Given transitions are

As given, $\lambda_{3}>\lambda_{2}>\lambda_{1}$
$\Rightarrow f_{3} < f_{3} < f_{1}$
Hence, $\lambda_{1}$ corresponds to $n = 3$ to $n = 1$ transition,
$\lambda_{2} $ corresponds to $ n = 2$ to $n = 1$ transition
and $\lambda_{3}$ corresponds to $n = 3$ to $n = 2$ transition.
Clearly, $E_{3 \rightarrow 1}=E_{3 \rightarrow2} +E_{2 \rightarrow1}$
$\frac{hc}{\lambda_{1}}=\frac{hc}{\lambda_{1} }+\frac{hc}{\lambda_{2} }$
$\frac{1}{\lambda_{1} }=\frac{1}{\lambda_{3} }+\frac{1}{\lambda _{2}} $