Modern Physics Ques 68
- 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) $388.9 $ $nm$
(c) $642.7 $ $nm$
(d) $488.9 $ $nm$
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Answer:
Correct Answer: 68.(d)
Solution:
Formula:
Wavelength Corresponding To Spectral Lines:
- Expression for the energy of the hydrogenic electron states for atoms of atomic number $Z$ is given by
$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, } (m < 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}}] $
$\Rightarrow \frac{1}{\lambda} \propto [\frac{1}{m^{2}}-\frac{1}{n^{2}}] Z^{2}$
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} \quad …….(i) \end{aligned} $
For second case, transition is from $n=4$ to $n=2$, i.e. $m=2$ and $n=4$
$\therefore \frac{1}{\lambda} \propto [\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}] \quad Z^{2} \Rightarrow \frac{1}{\lambda} \propto [\frac{1}{4}-\frac{1}{16}] Z^{2}$
$\text { or } \frac{1}{\lambda} \propto \frac{3}{16} Z^{2} \quad …….(ii)$
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} $
BETA
