Q89 / 90 · JEE Main 2024 (06 Apr Shift 1) · Physics
Radius of a certain orbit of hydrogen atom is $8.48 Ã…$. If energy of electron in this orbit is $E / x$. then $x=$ _____
(Given $\mathrm{a}_0=0.529 Ã…, E=$ energy of electron in ground state).
Correct answer: 16
Solution
We know
$\begin{aligned}
& \mathrm{r}=0.529 \frac{\mathrm{n}^2}{\mathrm{Z}} \Rightarrow 8.48=0.529 \frac{\mathrm{n}^2}{1} \\
& \mathrm{n}^2=16 \Rightarrow \mathrm{n}=4
\end{aligned}$
We know $\mathrm{E} \propto \frac{1}{\mathrm{n}^2}$ $\begin{aligned} & E_{n^{\text {th }}}=\frac{E}{16} \\ & x=16 \end{aligned}$
We know $\mathrm{E} \propto \frac{1}{\mathrm{n}^2}$ $\begin{aligned} & E_{n^{\text {th }}}=\frac{E}{16} \\ & x=16 \end{aligned}$
Hence, the correct answer is 16.