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(A) According to Bohr's model of the hydrogen atom:$(A)$ The speed of revolution of an electron in the $n^{th}$ orbit is given by $v_n = \frac{1}{4 \p

Vedclass · Accepted Answer

$A-i, B-iii, C-ii, D-iv$

Vedclass · Answer

(A) According to Bohr's model of the hydrogen atom:$(A)$ The speed of revolution of an electron in the $n^{th}$ orbit is given by $v_n = \frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$. Thus,$A \rightarrow i$.$(B)$ The kinetic energy of an electron in the $n^{th}$ orbit is $K.E. = \frac{1}{2} m v_n^2 = \left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$. Thus,$B \rightarrow iii$.$(C)$ The total energy of an electron in the $n^{th}$ orbit is $E_n = -K.E. = -\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$. Thus,$C \rightarrow ii$.$(D)$ The frequency of revolution is $f = \frac{v_n}{2 \pi r_n} = \left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$. Thus,$D \rightarrow iv$.Therefore,the correct matching is $A-i, B-iii, C-ii, D-iv$.

$A$. Speed of revolution of electron	$i$. $\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$
$B$. Kinetic energy	$ii$. $-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$C$. Total energy	$iii$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$D$. Frequency	$iv$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$

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