In a hydrogen atom, the electron is raised from its innermost orbit to an orbit of nine times the radius. The ratio of its period of revolution in the new orbit to that in the innermost orbit is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 27
In the Bohr model, the radius of the nth orbit of hydrogen is \( r_n=n^{2}r_1 \). The speed of the electron in it follows from Bohr’s second postulate, \( mv_nr_n=nh/2\pi \). The radius of the new orbit fixes its orbit number.
\( \begin{aligned} n^{2} &= \frac{r_n}{r_1} \\ &= 9 \end{aligned} \)
\( n=3 \)
The quantisation condition gives the speed in the nth orbit:
\( \displaystyle v_n=\frac{nh}{2\pi m r_n} \)
Since \( r_n \) grows as \( n^{2} \), the speed varies as \( 1/n \). The period of revolution is then:
\( \begin{aligned} T_n &= \frac{2\pi r_n}{v_n} \\ &\propto n^{2}\times n \\ &\propto n^{3} \end{aligned} \)
Therefore, the ratio of the periods in the third and the first orbits is:
\( \begin{aligned} \frac{T_3}{T_1} &= 3^{3} \\ &= 27 \end{aligned} \)