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In a hydrogen atom, the energy of the first excited state is $-3.4 \mathrm{eV}$. Then, find out the K.E. of the same orbit of the hydrogen atom.
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Verified Answer
The correct answer is:
$+3.4 \mathrm{eV}$
For hydrogen atom,
The kinetic energy is equal to the negative of the total energy. And the potential energy is equal to the twice of the total energy.
The first excited state energy of orbital $=-3.4 \mathrm{eV}$ and the kinetic energy of same orbital $=-(-3.4 \mathrm{eV})$ $=3.4 \mathrm{eV}$
Related Theory
In search of its lowest energy configuration the Hydrogen atom in the first excited state will de-excite to the ground-state, by emitting a photon of energy $\left(E_2-E_\nu\right)$.
Caution
The energy in an excited state is nothing but kinetic energy.
The kinetic energy is equal to the negative of the total energy. And the potential energy is equal to the twice of the total energy.
The first excited state energy of orbital $=-3.4 \mathrm{eV}$ and the kinetic energy of same orbital $=-(-3.4 \mathrm{eV})$ $=3.4 \mathrm{eV}$
Related Theory
In search of its lowest energy configuration the Hydrogen atom in the first excited state will de-excite to the ground-state, by emitting a photon of energy $\left(E_2-E_\nu\right)$.
Caution
The energy in an excited state is nothing but kinetic energy.
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