(a)(i) State the principal factor that determines the relative stability of a radioactive nucleus.
(ii) Arrange the following radioactive nucleus in decreasing order of stability. Justify your answer: X,W and Y:
\(^40{X}_20\) \(^920{Y}_36\) and \(^95{Z}_42\)
(b)(i) Explain the term ionization potential.
(ii)
The diagram above illustrates energy levels in the hydrogen atom. E, is the energy of the E\(_0\) ground state.
(i) When an electron makes a transition from level n = 3 to level n = 1, it emits a photon of wavelength 1.02x 10\(^{-7}\)m. Calculate E\(_0\).
(ii) Calculate the ionization potential of the hydrogen atom.
(c)(i) Explain the statement, the work function of sodium is 2.0 eV. (ii) Light of wavelength 160 mm is shone on the surface of a sodium metal of work function 2.0 eV. Determine whether photoelectrons will be emitted. [h = 6.6 x 10\(^{-34}\) Js, e = 3.0 x 10\(^{8}\)m/s, I eV = 1.6 x 10\(^{-19}\) J]
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Discussions (9)
(a)i. The principal factor that determines the relative stability of a radioactive nucleus is neutrons-proton ratio
(ii). Neutron—proton ratio = neutron/proton
For X
A = 40, Z = 20, n = 40 — 20 = 20
Neutron—proton ratio = 20/20 = 1
For W
A = 92, Z = 36, n = 92 — 36 = 56
Neutron—proton ratio = 56/36 = 1.56
For Y
A = 95, Z = 45, n = 95 — 45 = 50
Neutron—proton ratio = 50/45 = 1.11
Therefore, the other of decreasing stability is X, Y, W.
NOTE:
A = mass number, Z = atomic number, n = proton number.
An atom tends to be more stable if the number of proton is closer to the number of neutron.
(b)i. Ionization potential is the minimum energy required to remove (liberate) an electron completely from an atom. It is measured in electron-volts (eV).
(ii). When an electron jumps from one energy level to another, it emits or absorbs energy, E.
The energy E is given as;
E = hf or E = hc/λ
Therefore, E = (6.6 x 10⁻³⁴ x 3.0 x 10⁸)/1.02 x 10⁻⁷ = 1.94 x 10⁻¹⁸ J
In electron-volts (eV)
E = (1.94 x 10⁻¹⁸)/1.6 × 10⁻¹⁹ = 12.125 eV
Now
E = E₃ — E₀
12.125 = -1.50 — E₀
E₀ = -1.50 — 12 = -13.625 eV
(ii) ionization potential = E&—E₀ = 0 — (-13.625) = 13.625 eV
(c)i. It means that the minimum energy required to remove an electron from the surface of sodium metal is 2.0 eV.
(ii). Energy of illuminated light, E is given as;
E = hc/λ = (6.6 x 10⁻³⁴ x 3.0 x 10⁸)/160 x 10⁻⁹ = 1.2375 x 10⁻¹⁸ J
In eV,
E = 1.2375 x 10⁻¹⁸ / 1.6 x 10⁻¹⁹ = 7.73 eV.
Since the energy of the emitted photons (light), 7.73 eV is greater than the work function of sodium, 2.0 eV, the photoelectrons will be emitted.
(a)i. The principal factor that determines the relative stability of a radioactive nucleus is neutrons-proton ratio
(ii). Neutron—proton ratio = neutron/proton
For X
A = 40, Z = 20, n = 40 — 20 = 20
Neutron—proton ratio = 20/20 = 1
For W
A = 92, Z = 36, n = 92 — 36 = 56
Neutron—proton ratio = 56/36 = 1.56
For Y
A = 95, Z = 45, n = 95 — 45 = 50
Neutron—proton ratio = 50/45 = 1.11
Therefore, the other of decreasing stability is X, Y, W.
NOTE:
A = mass number, Z = atomic number, n = proton number.
An atom tends to be more stable if the number of proton is closer to the number of neutron.
(b)i. Ionization potential is the minimum energy required to remove (liberate) an electron completely from an atom. It is measured in electron-volts (eV).
(ii). When an electron jumps from one energy level to another, it emits or absorbs energy, E.
The energy E is given as;
E = hf or E = hc/λ
Therefore, E = (6.6 x 10⁻³⁴ x 3.0 x 10⁸)/1.02 x 10⁻⁷ = 1.94 x 10⁻¹⁸ J
In electron-volts (eV)
E = (1.94 x 10⁻¹⁸)/1.6 × 10⁻¹⁹ = 12.125 eV
Now
E = E₃ — E₀
12.125 = -1.50 — E₀
E₀ = -1.50 — 12 = -13.625 eV
(ii) ionization potential = E&—E₀ = 0 — (-13.625) = 13.625 eV
(c)i. It means that the minimum energy required to remove an electron from the surface of sodium metal is 2.0 eV.
(ii). Energy of illuminated light, E is given as;
E = hc/λ = (6.6 x 10⁻³⁴ x 3.0 x 10⁸)/160 x 10⁻⁹ = 1.2375 x 10⁻¹⁸ J
In eV,
E = 1.2375 x 10⁻¹⁸ / 1.6 x 10⁻¹⁹ = 7.73 eV.
Since the energy of the emitted photons (light), 7.73 eV is greater than the work function of sodium, 2.0 eV, the photoelectrons will be emitted.
