Atoms And Nuclei Question 51
Question: The Rydberg constant R for hydrogen is [MP PMT/PET 1998]
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}} $
