Question

Due to transitions among its first three energy levels, hydrogenic atom emits radiation at three discrete wavelengths ${\lambda }_1,{\lambda }_2$, and ${\lambda }_3({\lambda }_1<{\lambda }_2<{\lambda }_3)$. Then.

• A. ${\lambda }_1={\lambda }_2+{\lambda }_3$
• B. ${\lambda }_1+{\lambda }_2={\lambda }_3$
• C. $1/{\lambda }_1+1/{\lambda }_2=1/{\lambda }_3$
• D. $1/{\lambda }_1=1/{\lambda }_2+1/{\lambda }_3$

Answer 1

Answer: (C)

$1/{\lambda }_1+1/{\lambda }_2=1/{\lambda }_3$

The transitions from higher energy levels to the lower energy levels are shown in the figure.

From figure, $E_3=E_2+E_1$

Or $\frac{hc}{{\lambda }_3}=\frac{hc}{{\lambda }_2}+\frac{hc}{{\lambda }_1}$

$⟹$ $\frac{1}{{\lambda }_3}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$