atoms Question 21
Question: Q. 15. The short wavelength limit for the Lyman series of the hydrogen is $913.4 \AA$. Calculate the short wavelength limit for Balmer series of hydrogen spectrum.
A [CBSE SQP 2014]
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Solution:
Ans. Rydberg formula for the wavelengths of spectral lines in hydrogen spectrum is
$$ \frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) $$
The short wavelength limit $\lambda_{L}$ for the Lyman series would be
$$ \begin{array}{rlrl} \frac{1}{\lambda_{L}} & =R\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right)=R \ & \therefore & R & =\frac{1}{913.4 \AA} \end{array} $$
$\therefore$ The short wavelength limit $\lambda_{B}$ for the Balmer series, would be
$$ \begin{aligned} \frac{1}{\lambda_{B}} & =R\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)=\frac{R}{4} \ \therefore \quad \lambda_{B} & =\frac{4}{R}=4 \times 913.4 \AA \ & =3653.6 \AA \end{aligned} $$
[CBSE Marking Scheme 2014]
