Q. In two hydrogen atoms X and Y the electrons move around the nucleus in circular orbits of radius R and 4R respectively. The ratio of the time taken by them to compute one revolution is

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A

1 : 4

B

1 : 2

C

1 : 8

D

2 : 1

Solution:

$\text{r} \propto \frac{\left(\text{n}\right)^{\text{2}}}{\text{z}} \, \text{(z} \, \text{=} \, \text{1} \, \text{for} \, \text{H-atom)}$
$\frac{\left(\text{r}\right)_{\text{x}}}{\left(\text{r}\right)_{\text{y}}} = \left(\frac{\left(\text{n}\right)_{\text{x}}}{\left(\text{n}\right)_{\text{y}}}\right)^{2} \Rightarrow \frac{\left(\text{n}\right)_{\text{x}}}{\left(\text{n}\right)_{\text{y}}} = \frac{1}{2}$
$\text{T} \, \text{=} \, \frac{\text{2\pi r}}{\text{v}}$
$\text{T} \, \propto \, \frac{\text{n}^{\text{3}}}{\text{z}^{\text{2}}}$
$\frac{\left(\text{T}\right)_{\text{x}}}{\left(\text{T}\right)_{\text{y}}} = \left(\frac{\left(\text{n}\right)_{\text{x}}}{\left(\text{n}\right)_{\text{y}}}\right)^{3} = 1 : 8$

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