Search any question & find its solution
Question:
Answered & Verified by Expert
If the radius of electron orbit in the excited state of hydrogen atom is $476.1 \mathrm{pm}$, the energy of electron in that excited state in $\mathrm{J}$ is (Radius and energy of electron in the first orbit of hydrogen atom are $52.9 \mathrm{pm}$ and $-2.18 \times 10^{-18} \mathrm{~J}$ respectively)
Options:
Solution:
1275 Upvotes
Verified Answer
The correct answer is:
$-2.42 \times 10^{-19}$
Given : Radius of excited state of hydrogen
$$
\begin{aligned}
& \text { atom }=476 \cdot 1 \mathrm{PM} \\
& \qquad \begin{aligned}
E_1 & =2.18 \times 10^{-8} \mathrm{~J} / \text { atom } \\
E_n & =-2.18 \times 10^{-18} \cdot \frac{Z^2}{n^2} \\
r_1 & =529 \mathrm{pm} \\
r_n & =\frac{n^2 \times 527}{Z} ; 476.1=\frac{n^2 \times 52.7}{Z} \\
\therefore n^2=9 & \text { and } E_n=E_3=-\frac{2.18 Z^2 \times 10^{-18} \mathrm{~J}}{9} \\
& =-2.42 \times 10^{-19} \mathrm{~J}
\end{aligned}
\end{aligned}
$$
$$
\begin{aligned}
& \text { atom }=476 \cdot 1 \mathrm{PM} \\
& \qquad \begin{aligned}
E_1 & =2.18 \times 10^{-8} \mathrm{~J} / \text { atom } \\
E_n & =-2.18 \times 10^{-18} \cdot \frac{Z^2}{n^2} \\
r_1 & =529 \mathrm{pm} \\
r_n & =\frac{n^2 \times 527}{Z} ; 476.1=\frac{n^2 \times 52.7}{Z} \\
\therefore n^2=9 & \text { and } E_n=E_3=-\frac{2.18 Z^2 \times 10^{-18} \mathrm{~J}}{9} \\
& =-2.42 \times 10^{-19} \mathrm{~J}
\end{aligned}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.