Question

Find the ratio of the time period for $2^{nd}$ Bohr orbit of $H{e}^{+}$ and $4^{th}$ Bohr orbit of $L{i}^{2+}.$

• A. $32:9$
• B. $9:32$
• C. $9:16$
• D. None of these

Answer 1

The correct option is B $9:32$

As per Bohr's theory, $T_n\propto \frac{n^3}{Z^2}$
where, $T_n$ = Time period of revolution
n = orbit number and
Z = atomic number of element
For $H{e}^{+}$, $Z=2,n=2$
$T_{H{e}^{+}}=K\times \frac{2^3}{2^2}$
where, K is the proportionality constant
Similarly, for $L{i}^{2+}$, $Z=3,n=4$
$T_{L{i}^{2+}}=K\times \frac{4^3}{3^2}$
Taking the ratio of time period of $H{e}^{+}$ and $L{i}^{2+},$
$\therefore \frac{T_{2\left(H{e}^{+}\right)}}{T_{4\left(Li{}^{2+}\right)}}=\frac{\frac{2^3}{2^2}}{\frac{4^3}{3^2}}=\frac{9}{32}$