Question

In hydrogen spectrum, the shortest wavelength in Balmer series is the ${}^{\prime }\lambda {}^{\prime }$. The shortest wavelength in the Brackett series will be

• A. $2\lambda$
• B. $4\lambda$
• C. $9\lambda$
• D. $16\lambda$

Answer 1

The correct option is B $4\lambda$

For any series:
Shortest wavelength, for electron transit from $\infty$ to n
$\frac{1}{\lambda }=R\left(\frac{1}{n^2}\right)$

For Balmer series, $n=2$
$\frac{1}{\lambda }=R\left(\frac{1}{2^2}\right)=\frac{R}{4}$

Then for brackett series, $n=4$
$\frac{1}{{\lambda }_b}=R\left(\frac{1}{4^2}\right)=\frac{R}{16}$

Thus, ${\lambda }_b=4\lambda$