In hydrogen atom the kinetic energy of electron is 3.4eV. The distance of that electron from the nucleus is 1. 2.116A°2. 0.529A°3. 1.587A°4. 21.16A°

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Vikas TU · Answer

Hiii Here the KE = 3.4eV So, electron is in first excited state. n = 2 So, distance from the nucleus = 0.529 * n 2 /z => 2.116 A 0 so, optin A is correct.

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In hydrogen atom the kinetic energy of electron is 3.4eV. The distance of that electron from the nucleus is 1. 2.116A°2. 0.529A°3. 1.587A°4. 21.16A°

In hydrogen atom the kinetic energy of electron is 3.4eV. The distance of that electron from the nucleus is 1. 2.116A°2. 0.529A°3. 1.587A°4. 21.16A°

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In hydrogen atom the kinetic energy of electron is 3.4eV. The distance of that electron from the nucleus is 1. 2.116A°2. 0.529A°3. 1.587A°4. 21.16A°

In hydrogen atom the kinetic energy of electron is 3.4eV. The distance of that electron from the nucleus is 1. 2.116A°2. 0.529A°3. 1.587A°4. 21.16A°

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