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  • Ratio of orbital frequency of electron of hydrogen in 3rd and 2nd orbital is:Option: 1 <img alt="\frac{27}{8}" src="https://lh4.googleusercon

Ratio of orbital frequency of electron of hydrogen in 3rd and 2nd orbital is:

Option: 1

\frac{27}{8}


Option: 2

\frac{8}{27}


Option: 3

\frac{4}{9}


Option: 4

\frac{9}{4}


Answers (1)

best_answer

Orbital frequency : 

f=\frac{mz^{2}e^{4}}{4\epsilon _{2}^{0}n^{3}h^{3}}

f\alpha \frac{z^{2}}{n^{3}}

m & e are mass and charge of electron 

h = planck's constant

z = atomic number

n = principal quantum number

 

As we have learnt,

The orbital frequency in 3rd orbit is given as:

\mathrm{f_{3}\, =\, \frac{mz^{2}e^{4}}{4\epsilon _{2}^{0}n^{3}h^{3}}\, =\, \frac{mz^{2}e^{4}}{4\epsilon _{2}^{0}h^{3}}\times \frac{1}{27}}

And, the orbital frequency in 2nd orbit is given as:

\mathrm{f_{2}\, =\, \frac{mz^{2}e^{4}}{4\epsilon _{2}^{0}n^{3}h^{3}}\, =\, \frac{mz^{2}e^{4}}{4\epsilon _{2}^{0}h^{3}}\times \frac{1}{8}}

Thus,

\begin{aligned} \mathrm{f}_3 / \mathrm{f}_2 & =(1 / 27) /(1 / 8) \\ & =8 / 27 \end{aligned}

Therefore, Option(2) is correct.

Posted by

Sanket Gandhi

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