In the Hydrogen atom, the energy of the first excited state is $- 3.4 e V$ . Then find out KE of the same orbit of the Hydrogen atom:

+ 3.4 eV

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+ 6.8 eV

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13.6 eV

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+ 13.6 eV

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Solution

The correct option is A + 3.4 eV
Total energy of electron, $E_{T} =$ Potential energy (PE) + Kinetic energy (KE)

For an electron revolving in a circular orbit of radius, r around a nucleus of H atom:
$P E = - K Z e^{2} / r$
and
$K E = K Z e^{2} / 2 r$

Thus, $E_{T} = (- K Z e^{2} / r) + (K Z e^{2} / 2 r) = - K Z e^{2} / 2 r = - K E$

Thus, for $E_{T} = - 3.4 e V$

$K E = + 3.4 e V$

Option A is correct.

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In the Hydrogen atom, the energy of the first excited state is −3.4eV. Then find out KE of the same orbit of the Hydrogen atom:

In the Hydrogen atom, the energy of the first excited state is $- 3.4 e V$ . Then find out KE of the same orbit of the Hydrogen atom: