SATHEE: Atoms And Nuclei Question 291

Atoms And Nuclei Question 291

Question: A hydrogen atom in its ground state absorbs 10.2 eV of energy The orbital angular momentum is increased by

Options:

A) $1.05 \times 10^{- 34} J - s$

B) $3.16 \times 10^{- 34} J - s$ [b] $E_{p} = \frac{2 K_{e x}}{{(x^{2} + \frac{b^{2}}{4})}^{3 / 2}}$ For maximum $E_{p}$ $\frac{d E_{p}}{d x} = 0$
$\Rightarrow x = \frac{b}{2 \sqrt{2}}$

C) $2.11 \times 10^{- 34} J - s$

D) $4.22 \times 10^{- 34} J - s$

Show Answer

Answer:

Correct Answer: A

Solution:

Electron after absorbing 10.2 eV energy goes to its first excited state (n = 2) from ground state (n = 1).
$∴$ Increase in momentum $= \frac{h}{2 π}$ $= \frac{6.6 \times 10^{- 34}}{6.28} = 1.05 \times 10^{- 34} J - s$

Atoms And Nuclei Question 291

Question: A hydrogen atom in its ground state absorbs 10.2 eV of energy The orbital angular momentum is increased by

Options:

Answer:

Solution:

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Atoms And Nuclei Question 291

Question: A hydrogen atom in its ground state absorbs 10.2 eV of energy The orbital angular momentum is increased by

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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