1. Ionization Energy Of Hydrogen Kj/mol
2. Ionization Energy Of Hydrogen
3. Ionization Energy Of Hydrogen Joules

1. When we learned about periodic trends, we learned about ionization energy. Just how much energy is required to remove an electron from an atom?
2. One mole of hydrogen atoms has an atomic weight of 1.00 gram, and the ionization energy is 1,312 kilojoules per mole of hydrogen. The ionization energy is a measure of the capability of an element to enter into chemical reactions requiring ion formation or donation of electrons.
3. The energy of an electron in Bohr’s orbit of Hydrogen atom is given by the expression: E n = 2 π 2 m e 4 Z 2 n 2 h 2 (4 π ϵ 0) 2 = − 13.6 Z 2 n 2 e V Since Z = 1 for hydrogen, the above equation can be further simplified. Now just plug in the required values.

The equation above solves to be -9.113 x 10 -8 meters, or – 91.13 nanometers, a match of known hydrogen ionization energy data. In fact, the calculated values for each of the remaining orbitals (displayed in red in the table below) exactly match the known ionization energies and wavelengths of hydrogen data, shown in italics in the table. Dissociation and Ionization of Hydrogen Now, the above formula can be used as a template for the reactions of interest for this write-up. The dissociation of molecular hydrogen and the ionization of atomic hydrogen:! H 2 '2H'2p 2e# Letting the initial concentration of H 2 be n 0, we can write down the concentration of all.

Hydrogenic atoms

Scaling rules

Problem:

Use the measured ionization energy of Helium (E_ionization = 24.6 eV) to calculate the energy of interaction of the two electrons in the ground state of the atom.

Solution:

• Concepts:
  Hydrogenic atoms
• Reasoning:
  If the electrons were not interacting with each other but only with the Z = 2 nucleus, the ionization energy of He would be that of a Hydrogenic atom with a Z = 2 nucleus. The difference this ionization energy and the measured ionization energy is equal to the interaction energy of the two electrons.
• Details of the calculation:
  For Hydrogenic atoms:
  E_I' = Z²E_I(μ'/μ) . If (μ'/μ) ~ 1, then E_I' = Z²13.6 eV.
  For non-interacting electrons and Z = 2 we have E_I' = 54.4 eV.
  (The binding energy is the negative of the ionization energy.)
  The interaction energy of the two electrons therefore is +29.8 eV.

Problem:

Consider an 'atom' of positronium made up of an electron and a positron.
Calculate the energies of the lowest three states of the system and briefly enumerate correction terms you have neglected.

Solution:

• Concepts:
  The hydrogenic atom
• Reasoning:
  Positronium, made up of an electron and a positron, is a hydrogenic atom.

• Details of the calculation:
  H₀ = p²/(2μ) - e²/r is the Hamiltonian for the relative motion of the two particles. r = r₁ - r₂ is the vector pointing from particle 2 to particle 1.
  μ = m_e-m_e+/(m_e- + m_e+) = m_e/2 is the reduced mass of the system. H₀ is symmetric under the interchange of the two particles. H₀ neglects relativistic effects.
  Eigenvalues of H₀: E_n = -me⁴/(2ħ²n²) = -13.6/(2n²) eV.
  E_n depends only on n.
  Lowest energy level: n = 1, l = 0.
  First excited state: n = 2, l = 0, or l = 1.
  Relativistic effects, and spin-orbit and spin-spin couplings have been neglected.

Problem:

The ground state wave function for a hydrogen like atom is
Φ₁₀₀(r) = (1/√π)(Z/a₀)^3/2 exp(-Zr/a₀),
where a₀ = ħ²/(μe²) and μ is the reduced mass, μ ~ m_e = mass of the electron.
(a) What is the ground state wave function of tritium?
(b) What is the ground state wave function of ³He⁺?
(c) An electron is in the ground state of tritium. A nuclear reaction instantaneously changes the nucleus to ³He⁺. Assume the beta particle and the neutrino are immediately removed from the system. Calculate the probability that the electron remains in the ground state of ³He⁺.

Problem:

The emission spectrum of hydrogenic Lithium ions is measured and the wavelength of a series of emission lines are recorded. The longest line in that series has a wavelength of 450.25 nm. What is the wavelength of the shortest line in that series?

Solution:

• Concepts:
  The spectrum of hydrogenic atoms
• Reasoning:
  Longest wavelength in an emission series of hydrogenic ions:
  hc/λ_long = Z²*13.6 eV(1/n² - 1/(n+1)²).
• Details of the calculation:
  Z = 3 and hc = 1240 eV nm.
  λ_long = 450.25 nm = (10.13 nm) /(1/n² - 1/(n+1)². Solve for n, n = 4.
  For the shortest wavelength we have hc/λ_short = (Z²*13.6 eV)/n².
  λ_short = (10.13 nm)*16 = 162.09 nm.

Ionization Energy Of Hydrogen Kj/mol

Problem:

A certain species of ionized atoms produces an emission line spectrum according to the Bohr model, but the number of protons in the nucleus is unknown. A group of lines in the spectrum forms a series in which the shortest wavelength is 22.79 nm and the longest wavelength is 41.02 nm.
(a) Find the next-to-longest wavelength in the series of lines.
(b) Find the number of protons in the nucleus.

Solution:

• Concepts:
  Line spectrum of a hydrogenic atom
• Reasoning:
  E_n' - E_n = Z²*13.6 eV(1/n² - 1/n'²) = (hc/λ)(1/n² - 1/n'²).
• Details of the calculation:
  (a) 1/λ = (Z²*13.6/hc)(1/n² - 1/n'²) = C(1/n² - 1/n'²).
  1/λ_shortest = C/n², 1/λ_longest = C(1/n² - 1/(n + 1)²), λ_shortest/λ_longest= 1 - n²/(n + 1)² = 22.79/41.02.
  n²/(n + 1)² = 0.444, n = (n + 1)*0.666, n = 2.
  λ_shortest/λ_nl = 1 - n²/(n + 2)² = ¾. λ_nl = (4/3) 22.79 nm = 30.38 nm.
  (b) 1/λ_shortest = C/4, C = 0.1755/nm, Z = 4 is the number of protons in the nucleus.

Problem:

Consider a quasi-stable muonic atom composed of a muon in a 1s state and 9 electrons around a Ne nucleus (Z = 10).
(a) Calculate the approximate energy and radius of the lowest energy bound state of the muon. Comment on the effect of the outer electrons and of the finite size of the nucleus on the wave functions of this state.
(b) Calculate the approximate energy difference between the lowest energy electron states in the muonic atom and in ordinary neon.

Solution:

• Concepts:
  The hydrogenic atom
• Reasoning:
  To find the eigenfunctions and eigenvalues of the Hamiltonian of a hydrogenic atom we replace in the eigenfunctions of the Hamiltonian of the hydrogen atom a₀ by a₀' = ħ²/(μ'Ze²) = a₀(μ/μ')(1/Z), and in the eigenvalues of the Hamiltonian of the hydrogen atom we replace E_I by E_I' = μ'Z²e⁴/(2ħ²) = E_I(μ'/μ)Z².
  m_muon = 207 m_e. Therefore the average radius of the 1s state of the muon in Ne is approximately 1/200 times the average radius of the 1s state of the electron in Ne. For the muon the nuclear charge is hardly screened and we can use the hydrogenic atom formalism to solve for its energy.
  The muon, however screens the nuclear charge, and the outer electron of Ne move in the potential of a nucleus with charge Z - 1 = 9.
• Details of the calculation:
  (a) The reduced mass of the electron-Ne system is ~m_e and the reduced mass of the muon-Ne system is ~m_muon = 207 m_e. For the 1s state of the muon in Ne we have
  a₀' = a₀(μ/μ')(1/Z) = a₀(1/2070) = 25 fm,
  and E_I' = E_I(μ'/μ)Z² = E_I(207*100) = 0.28 MeV.
  The energy of the lowest energy bound state of the muon is -E_I'.
  Since the size of the nucleus is on the order of 10^-15m, the radius of the region in which the potential differs from a pure Coulomb potential is not completely negligible, and we may want to use perturbation theory to find correction terms for the energy and the wave function.
  (b) In ordinary Ne the energy of the lowest energy bound state of an electron is -13.6*100 eV.
  In Ne with a muon in a 1s state it is -13.6*81 eV.
  The absolute value of the difference is 258 eV.

Wave functions

Problem:

Certain nuclei can occasionally de-excite by internal conversion, which is a process whereby the excitation energy is transferred directly to one of the atomic electrons, causing it to be ejected from the atom. (This process competes with de-excitation by photon emission.) It is a reasonable assumption that the probability of this occurrence on a particular electron is directly proportional to the probability of that electron being at the nucleus. Of the n = 2 electrons, which are the most likely to undergo internal conversion and why? Estimate the ratio of conversion of 1s electrons to that of 2s electrons. Assume that the nuclear excitation energy is much greater than the ionization energy of the 1s electron.

Ionization Energy Of Hydrogen

Hydrogenic atoms wave functions:
R₁₀(r) = 2 (Z/a)^3/2 exp(-Zr/a),
R₂₀(r) = (Z/(2a))^3/2 (2 - Zr/a) exp(-Zr/(2a)),
R₂₁(r) = 3^-½(Z/(2a))^3/2(Zr/a) exp(-Zr/(2a)),
Y₀₀ = (4π)^-½, Y_1±1 = ∓(3/8π)^½sinθ exp(±iφ), Y₁₀ = (¾π)^½cosθ.

Solution:

• Concepts:
  Hydrogenic atoms, fundamental assumptions of Quantum Mechanics
• Reasoning:
  The probability that the electron is found at the nucleus is proportional to |ψ_nlm(r=0)|².

• Details of the calculation:
  ψ_2p(0) = 0, ψ_2s(0) ≠ 0. The 2s electron is more likely to undergo internal conversion than the 2p electron.
  Ratio: (1s internal conversion)/(2s internal conversion) = |ψ_1s(0)|²/|ψ_2s(0)|² = 8.

Ionization Energy Of Hydrogen Joules

Problem:

For hydrogen-like atoms, such as the alkali atoms, the screening effect of the 'closed-shell' electrons can be accounted for by considering the electron to move in the potential V(r) = -(e²/r)(1 + α/r), where α is a constant. Find the energy eigenvalues for this potential.

Solution:

• Concepts:
  Hydrogen-like atoms, the energy eigenfunctions and eigenvalues of the hydrogen atom
• Reasoning:
  The same equations have the same solutions.

• Details of the calculation:
  The radial equation for u(r) = rR(r) is
  [(∂²/∂r²) - l(l + 1)/r² + (2μ/ħ²)(e²/r)(1 + α/r) - k²_kl] u_kl(r) = 0,
  with k²_kl = -2μE_kl/ħ².
  We can write
  [(∂²/∂r²) - C_l/r² + (2μ/ħ²)(e²/r) - k²_kl] u_kl(r) = 0,
  with C_l = l(l + 1) - (2μe²α/ħ²).
  Changing to the variable ρ = r/a with a = ħ²/(μe²) we have
  [(∂²/∂ρ²) - C_l/ρ² + (2/ρ) - λ_kl²)] u_kl(ρ) = 0,
  with λ_kl² = -2E_kla/e².
  If we write C_l = l'(l' + 1), then this is the same equation we obtain for the hydrogen atom with l replaced by l'.

  For the hydrogen atom we have acceptable solutions if λ_kl = 1/(k + l), k = 1, 2, 3, ... .
  For the hydrogen atom l = 0, 1, 2, ... , so 1/(k + l) = 1/n, n = 1, 2, 3, ... .
  For our problem we have acceptable solutions if λ_kl = 1/(k + l').

  l'(l' + 1) = l(l + 1) - (2μe²α/ħ²) = l(l + 1) - 2α/a.
  l' = -½ + (¼ + l(l + 1) - 2α/a)^½ = -½ + ((l + ½)² - 2α/a)^½.
  Let n = k + l, k = n - l. Then λ_kl = 1/(n - l + l').
  λ_kl = 1/[n + ((l + ½)² - 2α/a)^½ - (l + ½)].
  Let D(l) = ((l + ½)² - 2α/a)^½ - (l + ½). Then

  E_kl = [e²/(2a)] [1/(n + D(l))²].
  The accidental degeneracy of the energy levels of hydrogen with respect to l is removed.

The ionization energy of the hydrogen atom is the energy required to remove its electron. Calculate the ionization energy of hydrogen atoms (J/atoms) for the electron:

in first orbital:
E = (-2.2x10^-18)/1^2 = -2.2x10^-18 J

in second orbital:
E = (-2.2x10^-18)/2^2 = -5.5x10^-19 J

Did I do these two right????

Seems right to me.

Z = atomic number, which is 1 for H.
n = valence

So, I would say yes.

I don't believe you need second answer. Hydrogen only has 1 electron. So there is no second orbital.

Hydrogen has a IE of 1312 kJ/mol only.

Which, if you use your first orbital number in J and convert to kJ you get:

-2.178 x 10^-18 J/atom x 6.022 x 10^23 atom/mol = 1.31 x 10^6

-1.31 x 10^6 J/mol x 1 kJ/1000J = roughly -1311.59 kJ/mol.

Not far off from 1312.