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

Answer: (B)

$n=2$

Answer 2

When an atom comes down from some higher energy level to the second energy level $(n_{1}=2)$ and $n_{2}=3,4,5,...$), then the lines of the spectrum are obtained in the visible part. $\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right)$, where $n=3,4,5, \ldots$ The shortest wavelength of the series corresponds to $n=\infty$ is $3646\, ?$