Modern Physics 1 Question 6
5. Taking the wavelength of first Balmer line in hydrogen spectrum $(n=3$ to $n=2)$ as $660 nm$, the wavelength of the $2^{\text {nd }}$ Balmer line $(n=4$ to $n=2)$ will be
(Main 2019, 9 April I)
(a) $889.2 nm($ b)
(b) $388.9 nm$
(c) $642.7 nm$
(d) $488.9 nm$
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Solution:
- Expression for the energy of the hydrogenic electron states for atoms of atomic number $Z$ is given by
$$ \begin{aligned} E & =h \nu=\frac{Z^{2} m e^{4}}{8 h^{2} E _0^{2}} \frac{1}{m^{2}}-\frac{1}{n^{2}} \quad \text { Here, } n \\ \text { or } \frac{h c}{\lambda} & =\frac{Z^{2} m e^{4}}{8 h^{2} E _0^{2}} \frac{1}{m^{2}}-\frac{1}{n^{2}} \Rightarrow \frac{1}{\lambda} \propto \frac{1}{m^{2}}-\frac{1}{n^{2}} Z^{2} \end{aligned} $$
For first case,
$$ \begin{aligned} & \lambda=660 nm, m=2 \text { and } n=3 \\ & \therefore \frac{1}{660 nm} \propto \frac{1}{(2)^{2}}-\frac{1}{(3)^{2}} Z^{2} \\ & \Rightarrow \frac{1}{660 nm} \propto \frac{1}{4}-\frac{1}{9} Z^{2} \text { or } \frac{5}{36} Z^{2} \end{aligned} $$
For second case, transition is from $n=4$ to $n=2$, i.e. $m=2$ and $n=4$
$$ \begin{array}{ll} \therefore & \frac{1}{\lambda} \propto \frac{1}{(2)^{2}}-\frac{1}{(4)^{2}} \quad Z^{2} \Rightarrow \frac{1}{\lambda} \propto \frac{1}{4}-\frac{1}{16} Z^{2} \\ \text { or } & \frac{1}{\lambda} \propto \frac{3}{16} Z^{2} \end{array} $$
From Eqs. (i) and (ii), we get
$$ \begin{aligned} \frac{\lambda}{660 nm} & =\frac{5}{36} \times \frac{16}{3} \\ \Rightarrow \quad \lambda & =\frac{80}{108} \times 660 nm=488.9 nm \end{aligned} $$
