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Question: Assume that the potential energy of an hydrogen atom in its ground state is zero. Then its energy in...

Assume that the potential energy of an hydrogen atom in its ground state is zero. Then its energy in the first excited state will be:
A.23.823.8 eVeV
B.27.527.5 eVeV
C.30.430.4 eVeV
D.34.834.8 eVeV

Explanation

Solution

Potential energy of atoms result from the movement of electrons in an orbit. In an excited state atoms move to their higher energy orbital. The potential energy increases as the electrons move far away from the nucleus.

Complete step by step answer:
Here it is given that the potential energy of the hydrogen atom in its ground state is zero.
Ground state is defined as the lowest energy of all the possible states whereas the excited state is defined as the energy having the highest value than the ground state.
P.E=2×13.6n2P.E=-2\times \dfrac{13.6}{{{n}^{2}}}
Where, P.EP.E is the potential energy
K.EK.E is the kinetic energy
Kinetic energy of hydrogen atom in ground state is 13.6n2\dfrac{13.6}{{{n}^{2}}} eVeV
n=1n=1 for hydrogen atom
Where, nn is principal quantum number
On substituting the value in above formula we get,
K.E=13.6(1)2K.E=\dfrac{13.6}{{{\left( 1 \right)}^{2}}}
K.E=13.6K.E=13.6 eVeV
Kinetic energy =13.6=13.6 eVeV
On substituting the value in potential energy we get,
In ground state, the potential energy of a hydrogen atom is 2×13.612-2\times \dfrac{13.6}{{{1}^{2}}} $$$$
Potential energy of a hydrogen atom in ground state is 27.2-27.2 eVeV
To make this potential energy equal to zero we need to add extra 27.227.2 eVeV energy
The total energy of a electron in first excited state of hydrogen atom is (13.6)n2\dfrac{\left( -13.6 \right)}{{{n}^{2}}}
T.E=(13.6)22T.E=\dfrac{\left( -13.6 \right)}{{{2}^{2}}}
T.E=3.4T.E=-3.4 eVeV
Where, T.ET.E is the total energy of the hydrogen atom and nn is the principal quantum number.
The total energy of an electron in the first excited state of a hydrogen atom is 3.4-3.4 eVeV
The energy level of a hydrogen atom is increased by 27.227.2 eVeV therefore, the total energy in the first excited state will be 3.4+27.2=23.8-3.4+27.2=23.8 eVeV
Hence, the correct option is (A)\left( A \right) that is 23.823.8 eVeV .

Additional information
Hydrogen atom is the simplest atom that has one proton and one electron. When an electron moves from energy level to another energy level then the energy of an electron also changes. Energy is required by the electron to move from state to another.

Note:
-Principal quantum number is a quantum number that is assigned to each electron in an atom that describes the electron state. It is denoted with the symbol nn.
-Potential energy is an energy associated with an object because of its position.