
An energy of 24.6eV is required to remove one of the electrons from a neutral helium atom. The energy (in eV) required to remove both the electrons from a neutral helium atom is
A. 38.2
B. 49.2
C. 51.8
D. 79.0
Answer
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Hint: A helium atom contains two electrons. If one of the electrons is removed then the helium atom becomes hydrogen like. In this situation, the Bohr has given a formula for energy of the electron in a hydrogen-like atom.
Formula used:
According to Bohr’s model of hydrogen atom, the energy of an electron in nth orbit around hydrogen atom is given as
${E_n} = - \dfrac{{13.6{Z^2}}}{{{n^2}}}eV{\text{ }}...\left( i \right)$
Here Z is the atomic number of the hydrogen like atom while n signifies the number of the orbit in which the electron is revolving.
Detailed step by step solution:
A Helium atom contains two electrons. The amount of energy required to remove the outermost electron from an atom is known as its ionization energy.
It is given that when one electron is removed from helium, the amount of energy required is given as
${E_1} = 24.6eV$
This is the ionization energy of the helium atom. Now we also want to remove the other electron from the helium atom. We notice that when one electron is removed from the helium then it becomes hydrogen-like which means that it contains only one electron like hydrogen atom.
Now using the Bohr model we can calculate the energy required to remove the second electron since now helium has become hydrogen like upon removal of one electron. Therefore, we can use equation (i) where Z = 2 for helium atom while n = 1 for ground state of the electron.
${E_2} = \dfrac{{13.6 \times 4}}{1} = 54.4eV$
Hence the total amount of energy required to remove the two electrons from helium atom is given as
$
E = {E_1} + {E_2} \\
= 24.6 + 54.4 \\
= 79eV \\
$
Hence, the correct answer is option D.
Note: The energy of electrons revolving around the nucleus is taken as negative because it is bounded due to attractive forces between electron and nucleus. The attractive forces are always taken as negative while the repulsive forces are taken as positive. The energy required to pull an electron out of the atom is positive.
Formula used:
According to Bohr’s model of hydrogen atom, the energy of an electron in nth orbit around hydrogen atom is given as
${E_n} = - \dfrac{{13.6{Z^2}}}{{{n^2}}}eV{\text{ }}...\left( i \right)$
Here Z is the atomic number of the hydrogen like atom while n signifies the number of the orbit in which the electron is revolving.
Detailed step by step solution:
A Helium atom contains two electrons. The amount of energy required to remove the outermost electron from an atom is known as its ionization energy.
It is given that when one electron is removed from helium, the amount of energy required is given as
${E_1} = 24.6eV$
This is the ionization energy of the helium atom. Now we also want to remove the other electron from the helium atom. We notice that when one electron is removed from the helium then it becomes hydrogen-like which means that it contains only one electron like hydrogen atom.
Now using the Bohr model we can calculate the energy required to remove the second electron since now helium has become hydrogen like upon removal of one electron. Therefore, we can use equation (i) where Z = 2 for helium atom while n = 1 for ground state of the electron.
${E_2} = \dfrac{{13.6 \times 4}}{1} = 54.4eV$
Hence the total amount of energy required to remove the two electrons from helium atom is given as
$
E = {E_1} + {E_2} \\
= 24.6 + 54.4 \\
= 79eV \\
$
Hence, the correct answer is option D.
Note: The energy of electrons revolving around the nucleus is taken as negative because it is bounded due to attractive forces between electron and nucleus. The attractive forces are always taken as negative while the repulsive forces are taken as positive. The energy required to pull an electron out of the atom is positive.
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