SATHEE: Atoms And Nuclei Question 110

Atoms And Nuclei Question 110

Question: An electron jumps from 5th orbit to 4th orbit of hydrogen atom. Taking the Rydberg constant as $10^{7}$ per metre. What will be the frequency of radiation emitted [Pb. PMT 2001]

Options:

A) $6.75 \times 10^{12}, H z$

B) $6.75 \times 10^{14}, H z$

C) $6.75 \times 10^{13}, H z$

D) None of these

Show Answer

Answer:

Correct Answer: C

Solution:

By using $ν = R C, [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$\Rightarrow ν = 10^{7} \times (3 \times 10^{8}), [\frac{1}{4^{2}} - \frac{1}{5^{2}}]$ = 6.75 ´ 1013 Hz

Atoms And Nuclei Question 110

Question: An electron jumps from 5th orbit to 4th orbit of hydrogen atom. Taking the Rydberg constant as $10^{7}$ per metre. What will be the frequency of radiation emitted [Pb. PMT 2001]

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

Atoms And Nuclei Question 110

Question: An electron jumps from 5th orbit to 4th orbit of hydrogen atom. Taking the Rydberg constant as 107 per metre. What will be the frequency of radiation emitted [Pb. PMT 2001]

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

Menu

Question: An electron jumps from 5th orbit to 4th orbit of hydrogen atom. Taking the Rydberg constant as $10^{7}$ per metre. What will be the frequency of radiation emitted [Pb. PMT 2001]