Atoms And Nuclei Question 428

Question: The Rydberg constant R for hydrogen is

Options:

A) $ R=-( \frac{1}{4\pi {\varepsilon_{0}}} ).\frac{2{{\pi }^{2}}me^{2}}{ch^{2}} $

B) $ R=( \frac{1}{4\pi {\varepsilon_{0}}} ).\frac{2{{\pi }^{2}}me^{4}}{ch^{2}} $

C) $ R={{( \frac{1}{4\pi {\varepsilon_{0}}} )}^{2}}.\frac{2{{\pi }^{2}}me^{4}}{c^{2}h^{2}} $

D) $ R={{( \frac{1}{4\pi {\varepsilon_{0}}} )}^{2}}.\frac{2{{\pi }^{2}}me^{4}}{ch^{3}} $

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Answer:

Correct Answer: D

Solution:

  • $ R=\frac{2{{\pi }^{2}}k^{2}e^{4}m}{ch^{3}}={{( \frac{1}{4\pi {\varepsilon_{0}}} )}^{2}}\frac{2{{\pi }^{2}}me^{4}}{ch^{3}} $


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