Question

If the radius of the first orbit of the hydrogen atom is $5.29\times {10}^{-11}\text{m},$ the radius of the second orbit will be

• A. $21.16\times {10}^{-11}\text{m}$
• B. $10.58\times {10}^{-11}\text{m}$
• C. $15.87\times {10}^{-11}\text{m}$
• D. $2.64\times {10}^{-11}\text{m}$

Answer 1

The correct option is A $21.16\times {10}^{-11}\text{m}$

Given, $r_1=5.29\times {10}^{-11}\text{m}$

The orbital velocity of revolving electron in the $n^{th}$ orbit is,

$r_n=0.53\left(\frac{n^2}{Z}\right)\mathring{A}$

$r_n\propto \frac{n^2}{Z}[\because Z=\text{same}]$

Let $r_1$ and $r_2$ be the orbital radius of the $1^{st}$ and $2^{nd}$ orbit respectively.

$\therefore \frac{r_2}{r_1}={\left(\frac{n_2}{n_1}\right)}^2={\left(\frac{2}{1}\right)}^2$

$\Rightarrow r_2=4\times 5.29\times {10}^{-11}=21.16\times {10}^{-11}\text{m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.

Why this question ? Key concept : The orbital velocity of revolving electron in the $n^{th}$ orbit is, $r_n=0.53\left(\frac{n^2}{Z}\right)\mathring{A}$ $\Rightarrow r_n\propto \frac{n^2}{Z}$