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}} $ orbit to find out the energy of the second electron of Helium. Then from that value, we can find out the required value of the ionization potential.
Formula used: The formula used to solve this question is given by
$ E = - 13.6\dfrac{{{Z^2}}}{{{n^2}}}{\text{eV}} $ , here
Complete step-by-step solution
We know that the energy of the $ {n^{th}} $ orbit around the nucleus of an element is given by
$ E = - 13.6\dfrac{{{Z^2}}}{{{n^2}}}{\text{eV}} $ ........................(1)
Now, we know that the atomic number of Helium atom is equal to $ 2 $ . So both of its electrons must be present in the first orbit around its nucleus. Therefore, substituting $ Z = 2 $ in the above formula, we get
$ E = - 13.6\dfrac{{{2^2}}}{{{1^2}}}{\text{eV}} $
$ E = - 54.4{\text{eV}} $ ........................(2)
We know that this is the potential energy of the electron (with respect to the infinity) due to the attractive nuclear force. It is also known as the binding energy.
We also know that the ionization potential of an electron is the minimum energy required to be supplied to remove it from the atom. In other words, this should be the minimum energy to be supplied to the electron so that the electron is separated from the nucleus by an infinite distance. So the final energy of the electron will become zero. Let the ionization potential for the second electron be $ I $ . Since the final energy of the electron is zero, so we have
$ I + E = 0 $
$ \Rightarrow I = - E $
Putting (2) we finally get
$ I = 54.4{\text{eV}} $ .
Thus, the ionization potential for the second He electron is equal to $ {\text{54}}{\text{.4eV}} $ .
Hence, the correct answer is option D.
Note:
The second electron of the Helium atom does not mean that it belongs to the second state. This number is mentioned just to confuse us. The final answer would have been the same even if the ionization potential for the first electron was asked in the question.
Formula used: The formula used to solve this question is given by
$ E = - 13.6\dfrac{{{Z^2}}}{{{n^2}}}{\text{eV}} $ , here
Complete step-by-step solution
We know that the energy of the $ {n^{th}} $ orbit around the nucleus of an element is given by
$ E = - 13.6\dfrac{{{Z^2}}}{{{n^2}}}{\text{eV}} $ ........................(1)
Now, we know that the atomic number of Helium atom is equal to $ 2 $ . So both of its electrons must be present in the first orbit around its nucleus. Therefore, substituting $ Z = 2 $ in the above formula, we get
$ E = - 13.6\dfrac{{{2^2}}}{{{1^2}}}{\text{eV}} $
$ E = - 54.4{\text{eV}} $ ........................(2)
We know that this is the potential energy of the electron (with respect to the infinity) due to the attractive nuclear force. It is also known as the binding energy.
We also know that the ionization potential of an electron is the minimum energy required to be supplied to remove it from the atom. In other words, this should be the minimum energy to be supplied to the electron so that the electron is separated from the nucleus by an infinite distance. So the final energy of the electron will become zero. Let the ionization potential for the second electron be $ I $ . Since the final energy of the electron is zero, so we have
$ I + E = 0 $
$ \Rightarrow I = - E $
Putting (2) we finally get
$ I = 54.4{\text{eV}} $ .
Thus, the ionization potential for the second He electron is equal to $ {\text{54}}{\text{.4eV}} $ .
Hence, the correct answer is option D.
Note:
The second electron of the Helium atom does not mean that it belongs to the second state. This number is mentioned just to confuse us. The final answer would have been the same even if the ionization potential for the first electron was asked in the question.
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