Question

For Balmer series that lies in the visible region, the shortest wavelength corresponds to quantum number :

• A. $n=1$
• B. $n=2$
• C. $n=3$
• D. $n=\infty$

Answer 1

The correct option is D $n=\infty$

When an atom comes down from some higher energy level to the second energy level $(n_1=2and{n}_2=3,4,5,\dots )$, then the lines of the spectrum are obtained in the visible part.

$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{n^2})$, where $n=3,4,5,\dots$

The shortest wavelength of the series corresponds to $n=\infty$ is $3646$ $\mathring{A}$.