Question

The number of revolutions/sec made by an electron in $II$ orbit is $8$ times of the number of revolution/sec made by electron in $n^{th}$ orbit. The value of $n$ is :

• A. $3$
• B. $6$
• C. $5$
• D. $4$

Answer 1

The correct option is D $4$

The number of revolutions per second by an electron in a shell (orbital frequency) is given by the expression:
$\text{Orbital frequency}=\frac{\text{velocity of electron}}{\text{circumference of the orbit}}=\frac{u}{2\pi r}=\frac{E_1}{h}\left[\frac{2}{n^3}\right]$......(1)
Here, $E_1$ is the energy of first shell and $n$ is the principal quantum number of the orbit.
For the $n^{th}$ orbital, let $x$ be the orbital frequency.
For second orbital, the orbital frequency will be $8x$
Hence, $\frac{{\nu }^{\prime }}{\nu }=(\frac{n}{n^{\prime }}{)}^3$
$\frac{8x}{x}=(\frac{n}{2}{)}^3$
$8=(\frac{n}{2}{)}^3$
$2=\frac{n}{2}$
$n=4$
Thus, the value of n is 4.