Question

Find the wavelength of the limiting line in the Lyman series of the hydrogen spectrum? $(R=1.097\times {10}^{-2}{\text{nm}}^{-1})$

• A. $94.21\text{nm}$
• B. $91.16\text{nm}$
• C. $911.6\text{nm}$
• D. $933.6\text{nm}$

Answer 1

The correct option is B $91.16\text{nm}$

The limiting line will be the shortest line in the Lyman series. It is observed when an electron drops from an orbit with $n=\infty$ to an orbit with $n=1$.

The wavelength of this line is given by:
$\frac{1}{\lambda }=R(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})$

Given:
$R=1.097\times {10}^{-2}{\text{nm}}^{-1}$
Substituting the values of n and R we get:
$\frac{1}{\lambda }=1.097\times {10}^{-2}\times (\frac{1}{1^2}-\frac{1}{{\infty }^2})⟹\lambda =\frac{100}{1.097}=91.16nm$