The ratio of the area of the first excited st | vedclass

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

Vedclass · Accepted Answer

$16: 1$

Vedclass · Answer

(D) The radius of the $n^{th}$ orbit of a hydrogen atom is given by $r_n \propto n^2$. Since the area $A$ of the orbit is $A = \pi r_n^2$,we have $A_n \propto (n^2)^2 = n^4$. For the ground state,$n_1 = 1$. For the first excited state,$n_2 = 2$. The ratio of the area of the first excited state $(A_2)$ to the ground state $(A_1)$ is: $\frac{A_2}{A_1} = \left(\frac{n_2}{n_1}\right)^4 = \left(\frac{2}{1}\right)^4 = 16$. Therefore,the ratio is $16: 1$.

The ratio of the area of the first excited state to the ground state orbit of a hydrogen atom is:

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