The ratio of areas within the electron orbits | vedclass

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Question

(D) The radius of the $n^{th}$ orbit in a hydrogen atom is given by $r_n = a_0 n^2$, where $a_0$ is the Bohr radius and $n$ is the principal quantum n

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(D) The radius of the $n^{th}$ orbit in a hydrogen atom is given by $r_n = a_0 n^2$, where $a_0$ is the Bohr radius and $n$ is the principal quantum number.The area $A$ of the orbit is given by $A = \pi r_n^2 = \pi (a_0 n^2)^2 = \pi a_0^2 n^4$.Thus, the area is proportional to the fourth power of the principal quantum number: $A_n \propto n^4$.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:$\frac{A_2}{A_1} = \left( \frac{n_2}{n_1} \right)^4 = \left( \frac{2}{1} \right)^4 = \frac{16}{1}$.Therefore, the ratio is $16 : 1$.

The ratio of areas within the electron orbits for the first excited state to the ground state for a hydrogen atom is (in $ : 1$)

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