Question

The ratio of time taken by an electron in revolution around the H-nucleus in $2$nd and $3$rd orbits is:

• A. $2:3$
• B. $4:8$
• C. $8:27$
• D. $27:8$

Answer 1

The correct option is A $2:3$

time is taken for an electron to complete one revolution $=2\pi r/v$

Substitute the value of v, we get

$t=2\pi r/nh\times 2\pi mr$

$=4\pi 2m{r}^2/nh$

and $r_n=a_0{n}^2$

$t_n$ = time taken in one revolution in $n^{th}$ orbit of H atom

so $t_n=4\pi 2m{a}_0{n}^2/nh=4\pi mn/h$

$\frac{t_2}{t_3}=\frac{4\pi m\times 2/h}{4\pi m\times 3/h}=\frac{2}{3}$

So, the ratio of time taken by an electron in 2nd and 3rd orbit is 2:3