Chemistry - Atomic Structure Question with Solution | TestHub

According to Bohr's theory, Total energy is $E_n=K_n+V_n$ Kinetic energy $=K_n=\frac{1}{8\pi {\epsilon }_0}\cdot \frac{Z{e}^2}{r}$ Potential energy $=V_n=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Z{e}^2}{r}$ $\therefore E_n=\frac{1}{8\pi {\epsilon }_0}\cdot \frac{Z{e}^2}{r}-\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Z{e}^2}{r}$ Radius of $n$th orbit $\left(r_n\right)=\frac{n^2{h}^2{\epsilon }_0}{\pi mZ{e}^2}$ $$E_n=-\frac{m{e}^4}{8{\epsilon }_0^2{h}^2}\times \frac{1}{n^2}$$ (A) $\frac{V_n}{K_n}=\frac{-\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Z{e}^2}{r}}{\frac{1}{8\pi {\epsilon }_0}\cdot \frac{Z{e}^2}{r}}=-2$ Hence, $(A)$ match with $(R)$. (B) $E_n\propto \frac{1}{n^2}$ or $E_n\propto \frac{1}{r_n}$ Radius of $n$th orbit $r_n\propto E_n^x$ $\therefore x=-1$ Hence, (B) match with (Q). (C) Angular momentum $=\frac{h}{2\pi }\sqrt{l(l+1)l}=0,1,2,\dots$ For the lower orbit $n=1$ $\therefore l=0$ and $m=0$ Hence, angular momentum of lowest orbit $=\frac{h}{2\pi }\sqrt{0(0H)}=0$ (C) match with (P) (D) $\frac{1}{r^n}\propto Z^y$ as $r_n\propto \frac{1}{Z}\therefore y=1$ Hence, (D) match with (S).