Question

What would be maximum wavelength for Brackett series of hydrogen-spectrum?

• A. $74483\mathring{A}$
• B. $22790\mathring{A}$
• C. $40519\mathring{A}$
• D. $18753\mathring{A}$

Answer 1

The correct option is C $40519\mathring{A}$

Wavelength: $\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})Z^2$ where, $R=109737.3$ $c{m}^{-1}$

For hydrogen spectrum, $Z=1$

For maximum wavelength of Brackett series, $n_1=4$ and $n_2=5$

$\therefore$ $\frac{1}{{\lambda }_m}=109737.3\times (\frac{1}{4^2}-\frac{1}{5^2})1^2$ $c{m}^{-1}$

$⟹$ ${\lambda }_m=0.00040519$ cm $=40519$ $A^o$