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Question

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:

A
+ 3.4 eV
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B
+ 6.8 eV
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C
13.6 eV
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D
+ 13.6 eV
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Solution

The correct option is A + 3.4 eV
Total energy of electron, ET= Potential energy (PE) + Kinetic energy (KE)

For an electron revolving in a circular orbit of radius, r around a nucleus of H atom:
PE=KZe2/r
and
KE=KZe2/2r

Thus, ET=(KZe2/r)+(KZe2/2r)=KZe2/2r=KE

Thus, for ET=3.4eV

KE=+3.4eV

Option A is correct.

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