Modern Physics Ques 24
- The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561$ $ \AA$. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is
(2011)
(a) $1215 $ $\AA$
(b) $1640 $ $\AA$
(c) $2430 $ $\AA$
(d) $4687 $ $\AA$
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Answer:
Correct Answer: 24.(a)
Solution:
Formula:
Wavelength Corresponding To Spectral Lines:
For hydrogen or hydrogen type atoms,
$ \frac{1}{\lambda}=R Z^{2} \left(\frac{1}{n _f^{2}}-\frac{1}{n _i^{2}}\right) $
In the transition from $n _i \rightarrow n _f$
$ \therefore \quad \lambda \propto \frac{1}{Z^{2} \left(\frac{1}{n _f^{2}}-\frac{1}{n _i^{2}}\right)} $
$\therefore \quad \frac{\lambda _2}{\lambda _1}=\frac{Z _1^{2} \left(\frac{1}{n _f^{2}}-\frac{1}{n _i^{2}}\right) _1}{Z _2^{2} \left(\frac{1}{n _f^{2}}-\frac{1}{n _i^{2}}\right) _2}$
$ \lambda _2=\frac{\lambda _1 Z _1^{2}\left(\frac{1}{n _f^{2}}-\frac{1}{n _i^{2}}\right) _1}{Z _2^{2} \left(\frac{1}{n _f^{2}}-\frac{1}{n _i^{2}}\right) _2} $
Substituting the values, we have
$ =\frac{(6561 \AA)(1)^{2} (\frac{1}{2^{2}}-\frac{1}{3^{2}})}{(2)^{2} (\frac{1}{2^{2}}-\frac{1}{4^{2}})}=1215 $ $\AA $
BETA
