Question

What is the ratio of time period in the second orbit of hydrogen atom to the third orbit of $H{e}^{+}$ ion?

• A. $\frac{8}{27}$
• B. $\frac{32}{27}$
• C. $\frac{27}{32}$
• D. $\frac{27}{8}$

Answer 1

The correct option is B $\frac{32}{27}$

$H{e}^{+}$ is hydrogen like atom.
So according to the Bohr's theory of hydrogen atom, we have $T\propto \frac{n^3}{Z^2}$
Time period in the second orbit of hydrogen atom, $T_1\propto \frac{2^3}{1^2}$
Time period in the third orbit of $H{e}^{+}$ ion, $T_2\propto \frac{3^3}{2^2}$
Thus, $\frac{T_1}{T_2}=\frac{2^5}{3^3}=\frac{32}{27}$

Approach to solve time period:
As we know, $T\propto \frac{r}{v}$
Since $E\propto \frac{Z^2}{n^2}$ also $E=\frac{1}{2}m{v}^2$

Hence, $v\propto \frac{Z}{n}..\left(i\right)$
From Bohr's angular momentum, $mvr=\frac{nh}{2\pi }$ or $r\propto \frac{n}{v}$ or $r\propto \frac{n^2}{Z}..\left(ii\right)$
From $e{q}^n\left(i\right)\&\left(ii\right)$, $T\propto \frac{n^3}{Z^2}$