Question

A line in Lyman series of hydrogen atoms has a wavelength of $121.6$nm. The initial energy level of that electron is:

• A. L
• B. M
• C. N
• D. O

Answer 1

The correct option is A L

$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]z^2$

$\Rightarrow \frac{1}{121.6\times {10}^{-9}}=1.097\times {10}^7[1-\frac{1}{n_2^2}]1$

$\Rightarrow 1-\frac{1}{n_2^2}=\frac{1}{121.6\times {10}^{-9}\times 1.097\times {10}^7}$

$=\frac{1}{1.33}=0.75$

$\Rightarrow 0.25=\frac{1}{n_2^2}$

$\Rightarrow n_2^2=4$

$\Rightarrow n_2=2$

$\Rightarrow n_2=n_1+1$

$⟶n_2=2$ corresponds to $L$ .