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Q.
The wavelength of $H _\alpha$ line of Balmer series is $X \mathring{A}$. What is the $X$ of $H _\beta$ line of Balmer series
Structure of Atom
Solution:
$H _\alpha$ line of Balmer series means first line of Balmer series
$n_1=2, n_2=3$
$\bar{v}=\frac{1}{\lambda_\alpha}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5 R}{36}$
$\therefore \lambda_\alpha=\frac{36}{5 R}=X$
$H _\beta$ line of Balmer series means, second line of Balmer series, $n_1=2, n_2=4$
$\bar{v}=\frac{1}{\lambda_\beta}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3 R}{16}$
$\therefore \lambda_\beta=\frac{16}{3 R}=X$
When $\frac{36}{3 R}=X$
Then $\frac{16}{3 R}=\frac{X \times 5 R \times 36}{36 \times 3 R}=\frac{80 X}{108} \mathring{A}$