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Q.
The first line of the Lyman series in a hydrogen spectrum has a wavelength of $1210 \mathring{A}$. The corresponding line of a hydrogen- like atom of $Z = 11$ is equal to
Atoms
Solution:
For first line of Lyman series $n_1 = 1, n_2 = 2$
$\therefore \frac{1}{\lambda} = Z^2 R \frac{3}{4}$
In the case of hydrogen atom,
$Z = 1, \frac{1}{\lambda} = R{\frac{3}{4}}$
For hydrogen-like atom,
$Z = 11, \frac{1}{\lambda'} = 121 R\frac{3}{4}$
$\Rightarrow \frac{\lambda'}{\lambda} = \frac{3R}{4} \times \frac{4}{121R \times 3}$
$\Rightarrow \lambda' = \frac{\lambda}{121}$
$= \frac{1210}{121} = 10\,\mathring{A}$