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, $E_{3\to 1}=E_{3\to 2}+E_{2\to 1}$
$\frac{hc}{{\lambda }_1}=\frac{hc}{{\lambda }_1}+\frac{hc}{{\lambda }_2}$
$\frac{1}{{\lambda }_1}=\frac{1}{{\lambda }_3}+\frac{1}{{\lambda }_2}$