A) \[{{T}_{n}}\propto \frac{1}{{{n}^{2}}}\]
B) \[{{T}_{n}}\]is independent of \[n;{{r}_{n}}\propto n\]
C) \[{{T}_{n}}\propto \frac{1}{n};{{r}_{n}}\propto n\]
D) \[{{T}_{n}}\propto \frac{1}{n}\text{and}\,{{r}_{n}}\propto {{n}^{2}}\]
Correct Answer: B
Solution :
[b]; Applying Bohr model to the given system, \[\frac{m{{v}^{2}}}{{{r}_{n}}}=\frac{k}{{{r}_{n}}}\] ...(i) and\[mv{{r}_{n}}=\frac{nh}{2\pi }\]or\[v=\frac{nh}{2\pi m{{r}_{n}}}\] Put in (i), \[\frac{m}{{{r}_{n}}}\times \frac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}{{m}^{2}}r_{n}^{2}}=\frac{k}{{{r}_{n}}}\] \[r_{n}^{2}=\frac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}mk}\] ?(ii) \[\therefore \]\[r_{n}^{2}\propto {{n}^{2}}\]or\[r_{n}^{{}}\propto n\] K.E. of the electron,\[{{T}_{n}}=\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}m\frac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}{{m}^{2}}r_{n}^{2}}=\frac{{{n}^{2}}{{h}^{2}}}{8{{\pi }^{2}}mr_{n}^{2}}\] Using (ii), we get \[{{T}_{n}}=\frac{{{n}^{2}}{{h}^{2}}4{{\pi }^{2}}mk}{8{{\pi }^{2}}m{{n}^{2}}{{h}^{2}}}=\frac{k}{2}\] \[\therefore \]\[{{T}_{n}}\]is independent of n.
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