SATHEE: Atoms And Nuclei Question 23

Atoms And Nuclei Question 23

Question: If the wavelength of the first line of the Balmer series of hydrogen is $6561 A A$ , the wavelength of the second line of the series should be [CPMT 1984; DPMT 2004]

Options:

A) $13122 A A$

B) $3280 A A$

C) $4860 A A$

D) $2187 A A$

Show Answer

Answer:

Correct Answer: C

Solution:

The wavelength of spectral line in Balmer series is given by

$\frac{1}{λ} = R, [\frac{1}{2^{2}} - \frac{1}{n^{2}}]$

For first line of Balmer series, n = 3 Þ $\frac{1}{λ_{1}} = R, [\frac{1}{2^{2}} - \frac{1}{3^{2}}] = \frac{5 R}{36}$ ;

For second line n = 4. Þ $\frac{1}{λ_{2}} = R, [\frac{1}{2^{2}} - \frac{1}{4^{2}}] = \frac{3 R}{16}$

$\frac{λ_{2}}{λ_{1}} = \frac{20}{27} \Rightarrow λ_{1} = \frac{20}{27} \times 6561 = 4860, \AA$

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Atoms And Nuclei Question 23

Question: If the wavelength of the first line of the Balmer series of hydrogen is 6561 AA , the wavelength of the second line of the series should be [CPMT 1984; DPMT 2004]

Options:

Answer:

Solution:

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Question: If the wavelength of the first line of the Balmer series of hydrogen is $6561 A A$ , the wavelength of the second line of the series should be [CPMT 1984; DPMT 2004]