The total energy of hydrogen atom when electron is in the first excited state is about $- 3.4 e V$ . The kinetic energy and potential energy of the electron in this state respectively are

3.4 e V - 3.4 e V

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- 3.4 e V, - 6.8 e V

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3.4 e V, - 6.8 e V

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6.8 e V, - 3.4 e V

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Solution

The correct option is C $3.4 e V, - 6.8 e V$
We know
$Total Energy = - Kinetic Energy = \frac{Potential Energy}{2}$
Given total energy $= - 3.4 e V$
Hence
Kinetic energy = - Total energy = $3.4 e V$
and
Potential energy = $2 \times$ Total energy = $- 6.8 e V$

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The total energy of hydrogen atom when electron is in the first excited state is about −3.4 eV. The kinetic energy and potential energy of the electron in this state respectively are

The correct option is C 3.4 eV,−6.8 eVWe know Total Energy=−Kinetic Energy=Potential Energy2 Given total energy =−3.4 eV Hence Kinetic energy = - Total energy = 3.4 eV and Potential energy = 2×Total energy = −6.8 eV

The total energy of hydrogen atom when electron is in the first excited state is about $- 3.4 e V$ . The kinetic energy and potential energy of the electron in this state respectively are

The correct option is C $3.4 e V, - 6.8 e V$
We know
$Total Energy = - Kinetic Energy = \frac{Potential Energy}{2}$
Given total energy $= - 3.4 e V$
Hence
Kinetic energy = - Total energy = $3.4 e V$
and
Potential energy = $2 \times$ Total energy = $- 6.8 e V$