Question

The wavelength of light emitted due to transition of electron from second orbit to first orbit in hydrogen atom is

• A. $6563\mathring{A}$
• B. $4102\mathring{A}$
• C. $4861\mathring{A}$
• D. $1215\mathring{A}$

Answer 1

Answer: (D)

From Bohr's theory, wavelength of radiations emitted in $H$-atom
$\frac{1}{\lambda }=R\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$
Here $n_1=2,n_2=1,R=1.097\times 10^7{m}^{-1}$
$\therefore \frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4}R$
$\Rightarrow \lambda =\frac{4}{3R}=\frac{4}{3\times 1.097\times 10^7}$
$=1.215\times 10^{-7}m$
$=1215\mathring{A}$