Atoms And Nuclei Question 424
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 $ \overset{o}{\mathop{A}}, $ is
Options:
A) Balmer series
B) Lyman series
C) Paschen series
D) Pfund series
Show Answer
Answer:
Correct Answer: C
Solution:
(c) $ \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)
