Atoms And Nuclei Question 270
Question: The acceleration of an electron in the first orbit of the hydrogen atom (z=1) is:
Options:
A) $ \frac{h^{2}}{{{\pi }^{2}}m^{2}r^{3}} $
B) $ \frac{h^{2}}{8{{\pi }^{2}}m^{2}r^{3}} $
C) $ \frac{h^{2}}{4{{\pi }^{2}}m^{2}r^{3}} $
D) $ \frac{h^{2}}{4\pi m^{2}r^{3}} $
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Answer:
Correct Answer: C
Solution:
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Speed of electron in first orbit (n=1) of hydrogen atom (z=1), $ v=\frac{e^{2}}{2{\varepsilon_{0}}h} $ $ r=\frac{h^{2}{\varepsilon_{0}}}{\pi me^{2}}\Rightarrow {\varepsilon_{0}}=\frac{r\pi me^{2}}{h^{2}} $
Acceleration of electron,$ \frac{v^{2}}{r}=\frac{e^{4}}{4\varepsilon_0^{2}h^{2}}\times \frac{\pi me^{2}}{h^{2}{\varepsilon_{0}}} $
$ \frac{e^{4}\times \pi me^{2}}{4h^{4}\varepsilon _{0}^{3}} $
Eliminating $ {\varepsilon_{0}} $ $ =\frac{e^{4}\pi me^{2}h^{6}}{4h^{4}r^{3}{{\pi }^{3}}m^{3}e^{6}}=\frac{h^{2}}{4{{\pi }^{2}}m^{2}r^{3}} $
