Question

The ratio of number of revolutions of electron per
second in 2nd and 3rd orbit of H-atom is​???

Answer 1

Answer:

The ratio of number of revolutions of electron per  second in 2nd and 3rd orbit of H-atom is 27:8.

Explanation:

Frequency = $\frac{1}{\text{Time period}}$

$f=\frac{1}{T}=\frac{v}{2\pi r}$

r = radius of the shell

v = velocity of an electron.

An if the an electron is in nth shell:

$f_n=\frac{v_n}{2\pi r_n}$

As we know that $r_n , v_n$ are given as:

$r_n=\frac{0.53\times 10^{10}\times n^2}{Z}$

$v_n=2.165\times 10^6\times \frac{Z}{n}$

For n=2 of H-atom.

$r_2={0.53\times 10^{10}\times 2^2}{1} m$

$v_2=2.165\times 10^6\times \frac{1}{2} m/s$

$f_2=\frac{2.165\times 10^6\times \frac{1}{2} m/s}{2\pi \frac{0.53\times 10^{10}\times 2^2}{1} m}$..(1)

For n=3 of H-atom.

$r_3={0.53\times 10^{10}\times 3^2}{1} m$

$v_3=2.165\times 10^6\times \frac{1}{3} m/s$

$f_3=\frac{2.165\times 10^6\times \frac{1}{3} m/s}{2\pi \frac{0.53\times 10^{10}\times 3^2}{1} m}$..(2)

The ratio of $\frac{f_2}{f_3}$:

$\frac{f_2}{f_3}=\frac{27}{8}$ (Using (1) & (2)

The ratio of number of revolutions of electron per  second in 2nd and 3rd orbit of H-atom is 27:8.

Answer 2

Explanation:

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