SATHEE: Atoms And Nuclei Question 65

Atoms And Nuclei Question 65

Question: Ionization energy of hydrogen is 13.6 eV. If $h = 6.6 \times 10^{- 34} J -$ sec, the value of R will be of the order of [RPMT 1997]

Options:

A) $10^{10} m^{- 1}$

B) $10^{7} m^{- 1}$

C) $10^{4} m^{- 1}$

D) $10^{- 7} m^{- 1}$

Show Answer

Answer:

Correct Answer: B

Solution:

$E = - R c h$ Þ $R = - \frac{E}{c h} = - \frac{13.6 \times 1.6 \times 10^{- 19}}{3 \times 10^{8} \times 6.6 \times 10^{- 34}}$ $= 1.098 \times 10^{7}, p e r, m$

Atoms And Nuclei Question 65

Question: Ionization energy of hydrogen is 13.6 eV. If $h = 6.6 \times 10^{- 34} J -$ sec, the value of R will be of the order of [RPMT 1997]

Options:

Answer:

Solution:

Engineering Preparation

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Atoms And Nuclei Question 65

Question: Ionization energy of hydrogen is 13.6 eV. If h=6.6×10−34J− sec, the value of R will be of the order of [RPMT 1997]

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: Ionization energy of hydrogen is 13.6 eV. If $h = 6.6 \times 10^{- 34} J -$ sec, the value of R will be of the order of [RPMT 1997]