Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore would exist as Hg 80+ . Use the equation in Problem 8.57 to calculate the energy required for the last ionization step, that is, Hg 79 + ( g ) → Hg 80 + ( g ) + e −
Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore would exist as Hg 80+ . Use the equation in Problem 8.57 to calculate the energy required for the last ionization step, that is, Hg 79 + ( g ) → Hg 80 + ( g ) + e −
Solution Summary: The author explains how the ionization energy of the Hg
Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore would exist as Hg80+. Use the equation in Problem 8.57 to calculate the energy required for the last ionization step, that is,
Hg
79
+
(
g
)
→
Hg
80
+
(
g
)
+
e
−
Definition Definition Substance that constitutes everything in the universe. Matter consists of atoms, which are composed of electrons, protons, and neutrons. Different atoms combine together to give rise to molecules that act as a foundation for all kinds of substances. There are five states of matter based on their energies of attraction: solid, liquid, gases, plasma, and BEC (Bose-Einstein condensates).
In the hydrogen atom, the electron is in its lowest energy state, n=1. The maximum electron energy that a hydrogen atom can have is 0 kJ/mole, at which point the electron would essentially be removed from the atom and it would become a H+ ion. How much energy does it take to ionize one hydrogen atom in kilojoules per mole and in Joules per atom? (Hint: calculate ΔE where εfinal is zero and εinitial is –1312.04 kJ/mol.)
The energy needed for the following process is 3.851 ×
104
4+
Be(g) → Be(g) + 4e¯
The first and third ionization energies of beryllium are 8.99 × 10² and 1.485 × 10³
kJ
mol
beryllium-that is, the energy required for the process:
kJ
mol
Second ionization energy of beryllium:
2+
Be (g) → Be(g) + e
A hydrogen-like ion is an ion containing only one electron. The energies of the electron in a hydrogen-like ion are given by
E₁₁ = -(2.180 × 10 1) (2²) ( = =))
- 18
n
2
n
kJ
mol
where En is the energy of the electron in the hydrogen-like ion, ʼn is the principal quantum number, and Z is the atomic number of the element. Round your
answer to 4 significant digits.
Note: Reference the Fundamental constants table for additional information.
U
kJ
mol'
□x10
respectively. Calculate the second ionization energy of
G
oo
區
olo
Ar
KJ
Let's say we have an imaginary element y:
First ionization energy = 771 kJ/mol
Second ionization energy = 2121 kJ/mol
Third ionization energy = 3801 kJ/mol
Fourth ionization energy = 30,001 kJ/mol
which of the following statements is incorrect?
Y could belong to Group IIIA.
The last ionization energy is much greater than the third ionization energy because Y3+ consists of a noble-gas core.
Y could belong to group VA
The third ionization energy is greater than the second ionization energy because Z2+ has a bigger charge than Y+.
Y could belong to Group IIIB.
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell