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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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2.116A°

0.529A°

1.587A°

21.16A°

answer is A.

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Detailed Solution

$E = - \frac{13.6}{n^{2}} e V 3.4 = - \frac{13.6}{n^{2}} e V n = 2 R a d i u s = 0.529 \times n^{2} R a d i u s = 0.529 \times 2^{2} = 2.11 A^{o}$

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