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Match the following List-I with List-II in connection with Bohr's atomic model.
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The correct answer is:
i iii ii iv
A) Speed of revolution of electron,
$=\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$
(B) Kinetic energy $=\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
(C) Total Energy $=-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
(D) Frequency $=\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$
Hence, $\mathrm{A} \rightarrow \mathrm{i}$; $\mathrm{B} \rightarrow \mathrm{iii}$; $\rightarrow$ ii; $\mathrm{D} \rightarrow$ iv.
$=\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$
(B) Kinetic energy $=\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
(C) Total Energy $=-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
(D) Frequency $=\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$
Hence, $\mathrm{A} \rightarrow \mathrm{i}$; $\mathrm{B} \rightarrow \mathrm{iii}$; $\rightarrow$ ii; $\mathrm{D} \rightarrow$ iv.
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