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In hydrogen atom the kinetic energy of electron is 3.4 eV . The distance of that electron from the nucleus

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a
2.116Å
b
0.529Å
c
1.587Å
d
21.16Å

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detailed solution

Correct option is A

The formula of energy is

En=-13.6eV/atom×Z2n2

For hydrogen atom Z =1.

Kinetic energy of electron K.E = 3.4eV

Then total energy of electron Etotal=-K.E

So, Etotal=-3.4eV

Then -3.4eV=-13.6eV×12n2; Then, n=2

If n = 2 , then the radius of orbit is

rn=0.529×n2A°

r2=0.529×4A°

r of 2nd orbit=2.116A°

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