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# The total energy of the electron in the hydrogen atom in the ground state is -13.6 eV. Which of the following is its kinetic energy in the first excited state?\begin{align} & \text{A}\text{. 13}\text{.6 eV} \\ & \text{B}\text{. 6}\text{.8 eV} \\ & \text{C}\text{. 3}\text{.4 eV} \\ & \text{D}\text{. 1}\text{.825 eV} \\ \end{align}

Last updated date: 20th Jun 2024
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Hint: The total energy of the electron in its zero excited state is given. Now we have to find the energy of the electron in the hydrogen atom in its first excited state. After finding the energy of the electron in a hydrogen atom we need to find the Kinetic energy of that electron in the first excited state.

The total energy of the electron in hydrogen atom in ground state is given as -13.6 eV,
That is,
E= -13.6 eV.
Now according to the question it is given that, what is the kinetic energy of the hydrogen atom in its first excited state, the energy that is given to us in the question is in ground state so in first excited state energy will be,
${{E}_{n}}=\dfrac{E}{{{\left( n+1 \right)}^{2}}}eV$
${{E}_{1}}=\dfrac{-13.6}{{{2}^{2}}}eV$ ,
${{E}_{1}}=-3.4eV$.
Now we know that,
K.E = |E|.
K.E = |-3.4 eV|
K.E = 3.4eV.

So, the correct answer is “Option C”.

Note:
In the last part of the answer we got the result of the energy for an electron in the first excited state as -3.4 eV but, the formula for the K.E had a modulus operator which means it will only show the magnitude irrespective of the sign. In the ground state the value of ‘n’ in the equation ${{E}_{n}}=\dfrac{E}{{{\left( n+1 \right)}^{2}}}eV$, but in the next case when it comes to the first excited state it became one so we can say n gradually increases with increase in the excited state.