According to Bohr's atomic theory :
(A) Kinetic energy of electron is $\propto \frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}}$
(B) The product of velocity (v) of electron and principal quantum number (n). 'vn' $\propto \mathrm{Z}^{2}$.
(C) Frequency of revolution of electron in an orbit is $\propto \frac{Z^{3}}{n^{3}}$.
(D) Coulombic force of attraction on the electron is $\propto \frac{Z^{3}}{n^{4}}$. Choose the most appropriate answer from the options given below:
Correct Option: , 2
(A) $\mathrm{KE}=-\mathrm{TE}=13.6 \times \frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}} \mathrm{eV}$
$=K \mathrm{E} \propto \frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}}$
(B) $V=2.188 \times 10^{6} \times \frac{\mathrm{z}}{\mathrm{n}} \mathrm{m} / \mathrm{sec}$
So, $\mathrm{Vn} \propto Z$
(C) Frequency $=\frac{\mathrm{V}}{2 \pi_{\mathrm{r}}}$
So, $\mathrm{F} \propto \frac{\mathrm{Z}^{2}}{\mathrm{n}^{3}}$
$\left[\therefore r \propto \frac{n^{2}}{z}\right.$ and $\left.\vee \propto \frac{z}{n}\right]$
(D) Force $\propto \frac{\mathrm{z}}{\mathrm{r}^{2}}$
So, $F \propto \frac{Z^{3}}{n^{4}}$
So, only statement (A) is correct
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