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15eV is given to ${{\text{e}}^{ - 1}}$ in ${4^{th}}$ orbit then find its final energy when it comes out of ${\text{H - }}$ atom.
(A) $14.15{\text{eV}}$
(B) $13.6{\text{eV}}$
(C) $12.08{\text{eV}}$
(D) $15.85{\text{eV}}$

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Last updated date: 17th Jun 2024
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Answer
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Hint: To solve this question, we need to use the formula for the energy of an electron in the ${n^{th}}$ state of an atom. The final energy of the electron will be equal to the sum of its initial energy and the energy provided externally.
Formula used: The formula used to solve this question is given by
${E_n} = - 13.6\dfrac{{{Z^2}}}{{{n^2}}}{\text{eV}}$, here $E$ is the total energy of an electron in the ${n^{th}}$ orbit of an atom having the atomic number of $Z$.

Complete step-by-step solution:
We know that the energy of an electron in the ${n^{th}}$ orbit around an atom is given by
$\Rightarrow$ ${E_n} = - 13.6\dfrac{{{Z^2}}}{{{n^2}}}{\text{eV}}$
Now, according to the question, the electron is revolving around the hydrogen atom. We know that the atomic number of the hydrogen is equal to $1$. Therefore we substitute $z = 1$ in the above equation to get
$\Rightarrow$ ${E_n} = - \dfrac{{13.6}}{{{n^2}}}{\text{eV}}$
Also, it is given that the electron is in the fourth orbit. So we substitute $n = 4$ to get
${E_n} = - \dfrac{{13.6}}{{{4^2}}}{\text{eV}}$
On solving we get
$\Rightarrow$ ${E_n} = - 0.85{\text{eV}}$........ (1)
Now, according to the question, $15{\text{eV}}$of energy is given to this electron. So its final energy will be given by
$\Rightarrow$ ${E_f} = 15{\text{eV + }}{{\text{E}}_n}$
From (1)
\[{E_f} = 15{\text{eV - 0}}{\text{.85eV}}\]
$\Rightarrow$ ${E_f} = 14.15{\text{eV}}$
Thus, the final energy of the electron with which it comes out of ${\text{H - }}$atom is equal to $14.15{\text{eV}}$.

Hence, the correct answer is option A.

Note: The magnitude of the energy of the electron in the ${n^{th}}$ orbit of an atom is equal to its ionization energy. The ionization energy is the minimum energy required to be supplied to an electron to make it free from the attractive force of the nucleus, or to make its total energy equal to zero.