SATHEE: Atoms And Nuclei Question 135

Atoms And Nuclei Question 135

Question: Taking Rydberg’s constant $R_{H} = 1.097 \times 10^{7} m$ first and second wavelength of Balmer series in hydrogen spectrum is [Pb. PMT 2004]

Options:

A) 2000 Å, 3000 Å

B) 1575 Å, 2960 Å

C) 6529 Å, 4280 Å

D) 6552 Å, 4863 Å

Show Answer

Answer:

Correct Answer: D

Solution:

$\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] .$ For first wavelength, $n_{1} = 2$ , $n_{2} = 3$
Þ $λ_{1} = 6563, \AA$ . For second wavelength, $n_{1} = 2$ , $n_{2} = 4$
Þ $λ_{2} = 4861, \AA$

Atoms And Nuclei Question 135

Question: Taking Rydberg’s constant $R_{H} = 1.097 \times 10^{7} m$ first and second wavelength of Balmer series in hydrogen spectrum is [Pb. PMT 2004]

Options:

Answer:

Solution:

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Atoms And Nuclei Question 135

Question: Taking Rydberg’s constant RH=1.097×107m first and second wavelength of Balmer series in hydrogen spectrum is [Pb. PMT 2004]

Options:

Answer:

Solution:

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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Question: Taking Rydberg’s constant $R_{H} = 1.097 \times 10^{7} m$ first and second wavelength of Balmer series in hydrogen spectrum is [Pb. PMT 2004]