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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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2.116Å

0.529Å

1.587Å

21.16Å

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detailed solution

Correct option is A

The formula of energy is

$E_{n} = \frac{- 13.6 eV / atom \times Z^{2}}{n^{2}}$

For hydrogen atom Z =1.

Kinetic energy of electron K.E = 3.4eV

Then total energy of electron $E_{total} = - K . E$

So, $E_{total} = - (3.4 eV)$

Then $- 3.4 eV = \frac{- 13.6 eV \times {(1)}^{2}}{n^{2}}; Then, n = 2$

If n = 2 , then the radius of orbit is

$r_{n} = 0.529 \times n^{2} A^{°}$

$r_{2} = 0.529 \times 4 A^{°}$

$∴ r of 2^{nd} orbit = 2.116 A^{°}$

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