SATHEE: Atoms And Nuclei Question 424

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 $\overset{o}{A},$ is

Options:

A) Balmer series

B) Lyman series

C) Paschen series

D) Pfund series

Show Answer

Answer:

Correct Answer: C

Solution:

(c) $\frac{1}{λ} = R, [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$\Rightarrow \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} = \frac{1}{R λ}$ $= \frac{1}{1.097 \times 10^{7} \times 18752 \times 10^{- 10}}$ $= 0.0486 = \frac{7}{144} .$ But $\frac{1}{3^{2}} - \frac{1}{4^{2}} = \frac{7}{144} \Rightarrow n_{1} = 3$ and n2 = 4 (Paschen series)

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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 Ao, is

Options:

Answer:

Solution:

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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 $\overset{o}{A},$ is