SATHEE: Atoms And Nuclei Question 143

Atoms And Nuclei Question 143

Question: The energy of electron in the nth orbit of hydrogen atom is expressed as $E_{n} = \frac{- 13.6}{n^{2}} e V$ . The shortest and longest wavelength of Lyman series will be [Pb. PET 2003]

Options:

A) 910 Å, 1213 Å

B) 5463 Å, 7858 Å

C) 1315 Å, 1530 Å

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

$\frac{1}{λ_{max}} = R, [\frac{1}{{(1)}^{2}} - \frac{1}{{(2)}^{2}}]$
Þ $λ_{max} = \frac{4}{3 R} \approx 1213, \AA$ and $\frac{1}{λ_{min}} = R, [\frac{1}{{(1)}^{2}} - \frac{1}{\infty}]$
Þ $λ_{min} = \frac{1}{R} \approx 910, \AA$

Atoms And Nuclei Question 143

Question: The energy of electron in the nth orbit of hydrogen atom is expressed as $E_{n} = \frac{- 13.6}{n^{2}} e V$ . The shortest and longest wavelength of Lyman series will be [Pb. PET 2003]

Options:

Answer:

Solution:

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Atoms And Nuclei Question 143

Question: The energy of electron in the nth orbit of hydrogen atom is expressed as En=−13.6n2eV . The shortest and longest wavelength of Lyman series will be [Pb. PET 2003]

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: The energy of electron in the nth orbit of hydrogen atom is expressed as $E_{n} = \frac{- 13.6}{n^{2}} e V$ . The shortest and longest wavelength of Lyman series will be [Pb. PET 2003]