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Q.
The ratio of areas of the electron orbits for the first excited state and the ground state for the hydrogen atom is
NTA AbhyasNTA Abhyas 2022
Solution:
The radius of the orbit of the electron in the $n$ th excited state
$ \, r_{e} \, =$ $\frac{n^{2} 4 \pi \epsilon _{0} h^{2}}{4 \pi ^{2} m Z e^{2}}$
For the first excited state
$ \, \, \, n=2 \, , \, Z=1$
$\because \, r ^{'}=$ $\frac{4 \epsilon _{0} h^{2}}{\pi m e^{2}}$
For the ground state of hydrogen atom
$ \, n=1 \, , \, Z=1$
$\because \, r ^{''}=$ $\frac{h^{2} \epsilon _{0}}{\pi m e^{2}}$
The ratio of radius
$ \, \frac{r ^{'}}{r ^{''}}=\frac{4}{1}$
The ratio of area of the electron orbit for hydrogen atom
$ \, \, \frac{\left(A \right)^{'}}{\left(A \right)^{''}}=\frac{4 \pi \left(r '\right)^{2}}{4 \pi \left(r ' '\right)^{2}}$
$ \, \frac{A ^{'}}{A ^{''}}=\frac{16}{1}$