Question

2. If the second Bohr orbit of H-atom has a radius of 2.21x10-10m. What is the radius of the sixth Bohr
orbit H-atom?
Options:
(a) 5.24x10-10m
(b) 8.84x10-10m
(c) 5.24x10-'1m
(d) 1.99x10-ºm
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According to Bobo​

Answer 1

Answer:

this questions answer is b

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Answer 2

Answer:

The radius of the sixth Bohr orbit of H-atom is equal to 1.99×10⁻⁹m.

Therefore, the option (d) is correct.

Explanation:

Given:

The radius of second Bohr orbit $(n_{1}=2)$ of H-atom $r_{1}=2.21*10^{-10}m$

Bohr's radius is given by :

$r=\frac{n^{2}h^{2}}{4\pi ^{2}me^{2}k}$                                                     .................(1)

where, n is principle quantum number

           h is plank's constant

           m is mass of an electron

            e is electron's electric charge .

From eq.(1) we can say that : r ∝ n²

Therefore $\frac{r_{1}}{r_{2}} =\frac{n_{1}^{2}}{n_{2}^{2}}$

   $r_{2}=\frac{r_{1}n_{2}^{2}}{n_{1}^{2}}$

Radius for the sixth Bohr's orbit of H-atom:

$r_{2}=\frac{(2.21*10^{-10}m)(6)^{2}}{(2)^{2}}$

$r_{2}=19.98*10^{-10}m$

$r_{2}=1.99*10^{-9}m$

Therefore the radius of sixth orbit of H-atom is 1.99×10⁻⁹m.