The ratio of the areas of the electron orbits | vedclass

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(D) For a hydrogen atom,the radius of the $n^{th}$ orbit is given by $r_n \propto n^2$.Since the area of the orbit $A_n = \pi r_n^2$,we have $A_n \pro

Vedclass · Accepted Answer

$81: 16$

Vedclass · Answer

(D) For a hydrogen atom,the radius of the $n^{th}$ orbit is given by $r_n \propto n^2$.Since the area of the orbit $A_n = \pi r_n^2$,we have $A_n \propto (n^2)^2 = n^4$.The first excited state corresponds to $n = 2$,and the second excited state corresponds to $n = 3$.Therefore,the ratio of the area of the second excited state $(A_3)$ to the first excited state $(A_2)$ is:$\frac{A_3}{A_2} = \left(\frac{3}{2}\right)^4 = \frac{81}{16}$.

The ratio of the areas of the electron orbits for the second excited state to the first excited state for the hydrogen atom is

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