SATHEE: Atoms - Result Question 15

Atoms - Result Question 15

16. In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

[2015 RS, 2013]

(a) $\frac{9}{4}$

(b) $\frac{27}{5}$

(d) $\frac{4}{9}$

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Answer:

Correct Answer: 16. (c)

Solution:

$\frac{1}{λ_{L}} = R [1 - \frac{1}{2^{2}}] = \frac{3 R}{4}$

For Balmer series $(3 \to 2)$

$\begin{aligned} \frac{1}{λ_{B}} = R [\frac{1}{4} - \frac{1}{9}] = \frac{5 R}{36} \\ \Rightarrow \frac{λ_{L}}{λ_{B}} = \frac{\frac{4}{3 R}}{\frac{36}{5 R}} = \frac{4}{36} (\frac{5}{3}) = \frac{5}{27} \end{aligned}$

The wavelength of spectral lines increases with the increases of order of the series.

$λ_{PFund} > λ_{Brackett} > λ_{Paschen} > λ_{Balmer} > λ_{Lymen}$

Atoms - Result Question 15

16. In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

Answer:

Engineering Preparation

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Atoms - Result Question 15

16. In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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