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Total energy of electron in an excited state of hydrogen atom is \( -3.4 \mathrm{eV} \). The kinetic and
potential energy of electron in this state
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potential energy of electron in this state
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The correct answer is:
\( \mathrm{K}=3.4 \mathrm{eV} \quad \mathrm{U}=-6.8 \mathrm{eV} \)
Given, total energy of electron $=-3.4 \mathrm{eV}$
Now, $\mid$ total energy $|=|$ kinetic energy $\mid$
Therefore, kinetic energy of electron in this excited state is $3.4 \mathrm{eV}$
Potential energy of electron $=2 \times$ Total energy
$=2 \times(-3.4 \mathrm{eV})=-6.8 \mathrm{eV}$
$(\because$ Total energy = Kinetic energy + Potential energy $\Rightarrow 2 \times$ Total energy = Potential energy )
Therefore, kinetic energy and potential energy of electron in this state is $3.4 \mathrm{eV}$ and $-6.8 \mathrm{eV}$, respectively
Now, $\mid$ total energy $|=|$ kinetic energy $\mid$
Therefore, kinetic energy of electron in this excited state is $3.4 \mathrm{eV}$
Potential energy of electron $=2 \times$ Total energy
$=2 \times(-3.4 \mathrm{eV})=-6.8 \mathrm{eV}$
$(\because$ Total energy = Kinetic energy + Potential energy $\Rightarrow 2 \times$ Total energy = Potential energy )
Therefore, kinetic energy and potential energy of electron in this state is $3.4 \mathrm{eV}$ and $-6.8 \mathrm{eV}$, respectively
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