Question

If 'x' is the radius of first Bohr's orbit of the
hydrogen atom. Then the ratio of radius of 2nd 4th and 6th orbit is​

Answer 1

$\large{ \underline{ \underline{ \bold{ \: Answer : \: \: \: }}}}$

$\to$ 1 : 4 : 9

$\large{ \underline{ \underline{ \bold{ \: Explanation : \: \: \: }}}}$

The radius of the $n^{th}$ orbit is

$r_n=\frac{n^2\times 52.9}{Z} \: \bold{(in \: \: pm) }$

where ,

$r_n$ = radius of $\bold{ {n}^{th} }$ orbit

n = number of orbit

Z = atomic number

Given that , x is the radius of first Bohr's orbit of the hydrogen atom

$\to$ $r_1=x=\frac{1^2\times 52.9}{1}</p><p>$

$\to$ $r_2 = \frac{2^2\times 52.9}{1}=4x$

$\to$ $r_4=\frac{4^2\times 52.9}{1}=16x</p><p>$

$\to$ $r_6 =\frac{6^2\times 52.9}{1}=36x</p><p>$

${\huge \star } \: \: r_2:r_4:r_6$

4x : 16x : 36x = 1 : 4 : 9

Therefore , the ratio of the radius of second , fourth and sixth orbit is 1: 4 : 9