Solving Nature of Atoms Homework: Part (a) ionization energy Z=2

In summary, part (a) is asking for the minimum energy (in electron volts) needed to remove an electron from the ground state of an ionized atom with Z=2. The answer is not 13.6 eV, which is the ionization energy for atomic hydrogen, but rather a different value for helium. This is because in ionization, the electron is completely removed from the atom instead of just being moved to an excited state. The equation E=-13.6*(Z^2/n^2) can be used to calculate the ionization energy, with Z=2 for helium and n=1 for the ground state.
  • #1
arod2812
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Homework Statement


(a) What is the minimum energy (in electron volts) that is required to remove the electron from the ground state of an ionized atom (Z=2)? (b) What is the ionization energy for this ion?


Homework Equations


After doing what I got the answer for part (b)... but I don't know what part (a) requires. I thought this number would be 13.6 eV, but it's not. What exactly is part (a) asking?


The Attempt at a Solution



I used the equation E= -13.6*(Z^2/n^2). I plugged in 2 for Z and 1 for n.
 
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  • #3
This gave me an answer of -54.4 eV. However, this negative value does not make sense for ionization energy.

Part (a) is asking for the minimum energy required to remove an electron from the ground state of an ionized atom with Z=2. This means that the atom already has a positive charge of +2, and the electron is being removed from this positively charged ion.

The equation you used, E= -13.6*(Z^2/n^2), is the Rydberg formula for calculating the energy levels of a hydrogen atom. However, it is not applicable for this problem since we are dealing with an ionized atom with Z=2, not a hydrogen atom.

To calculate the ionization energy, we need to use the equation E= -R*(Z^2/n^2), where R is the Rydberg constant (13.6 eV) and n is the principal quantum number of the electron being removed. Since we are dealing with an ionized atom, the principal quantum number of the electron in the ground state is n=1. Therefore, the ionization energy for this ion with Z=2 is E= -13.6*(2^2/1^2)= -13.6*4= -54.4 eV.

It is important to note that the negative sign in front of the ionization energy indicates that energy is required to remove the electron from the ion. This is because the electron is being removed from a positively charged ion, so energy must be added to overcome the attractive force between the positively charged ion and the negatively charged electron.

In conclusion, the minimum energy required to remove an electron from the ground state of an ionized atom with Z=2 is 54.4 eV, and the ionization energy for this ion is also 54.4 eV.
 

1. What is ionization energy?

Ionization energy is the amount of energy required to remove an electron from an atom in its gaseous state. It is measured in units of joules (J) or kilojoules per mole (kJ/mol).

2. How is ionization energy related to the nature of atoms?

The ionization energy of an atom is related to its electron configuration and the strength of the attraction between the nucleus and its electrons. It can provide information about the size of an atom and the ease with which it can form chemical bonds.

3. How is ionization energy calculated?

Ionization energy can be calculated by using the equation IE = -ΔE, where IE is the ionization energy and ΔE is the change in energy of the atom when an electron is removed. This can also be represented as the difference in energy between the neutral atom and the ion that is formed.

4. What does Z=2 represent in the ionization energy equation?

Z=2 represents the atomic number of the element, which is the number of protons in the nucleus of the atom. The atomic number determines the number of electrons in a neutral atom and can also provide information about the element's properties, such as its electron configuration.

5. How does ionization energy vary across the periodic table?

The ionization energy generally increases as you move across a period (row) in the periodic table, from left to right. This is because the number of protons in the nucleus increases, leading to a stronger attraction between the nucleus and the electrons. However, the ionization energy decreases as you move down a group (column) in the periodic table, as the electrons are further from the nucleus and are shielded by inner electrons.

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