Atoms And Nuclei Question 15
Question: Every series of hydrogen spectrum has an upper and lower limit in wavelength. The spectral series which has an upper limit of wavelength equal to 18752 Å is [MP PMT 1993]
Options:
A) Balmer series
B) Lyman series
C) Paschen series
D) Pfund series (Rydberg constant $ R=1.097\times 10^{7} $ per metre)
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Answer:
Correct Answer: C
Solution:
$ \frac{1}{\lambda }=R,[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} ] $
$ \Rightarrow \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}=\frac{1}{R\lambda } $
$ =\frac{1}{1.097\times 10^{7}\times 18752\times {{10}^{-10}}} $
$ =0.0486=\frac{7}{144}. $
But $ \frac{1}{3^{2}}-\frac{1}{4^{2}}=\frac{7}{144}\Rightarrow n_{1}=3 $ and n2 = 4 (Paschen series)
