Question

The ratio of the areas within the electron orbits for the first excited state, to the ground state, for the hydrogen atom is,

• A. $16:1$
• B. $18:1$
• C. $4:1$
• D. $2:1$

Answer 1

The correct option is A $16:1$

As we know that the radius of the electron in $n^{\text{th}}$ orbit is given as,
$r_n\propto n^2----\left(1\right)$
Also, the area of the $n^{\text{th}}$ orbit can be written as, $A\propto r_n^2---\left(2\right)$

From equations $\left(1\right)$ & $\left(2\right)$, we can write,
$A\propto n^4$

Let $A_0\text{and}{A}_1$ be the areas of the orbits of, ground state and first excited state respectively, then, $\frac{A_0}{A_1}=\frac{n_0^4}{n_1^4}$
Here, $n_0=1,n_1=2$

$\Rightarrow \frac{A_0}{A_1}=\frac{1}{16}$

or, $\frac{A_1}{A_0}=\frac{16}{1}$

Hence, $\left(A\right)$ is the correct answer.