Question

The maximum wavelength of Brackett series of hydrogen atom will be

• A. $35,890\mathring{A}$
• B. $14,440\mathring{A}$
• C. $62,160\mathring{A}$
• D. $40,400\mathring{A}$

Answer 1

The correct option is D $40,400\mathring{A}$

The maximum wavelength is formed when the transition occurs just from next orbit.
Now, for Brackett $n=4$
$\frac{hC}{\lambda }=13.6\times [\frac{1}{16}-\frac{1}{25}]$

$\Rightarrow \lambda =\frac{4.14\times {10}^{-15}\times 3\times {10}^8}{13.6\times 0.0225}$

$\Rightarrow \lambda =4.05882\times {10}^{-6}m$
$\Rightarrow \lambda =40,588\times {10}^{-10}m$

$\Rightarrow \lambda =40,558A^{\circ }\approx 40,400A^{\circ }$