(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:
\(\displaystyle {}^{40}_{20}X \quad {}^{920}_{36}Y \text{ and } {}^{95}_{42}Z\)
(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 \(1.02 \times 10^{-7}\,\text{m}\). 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. [\(h = 6.6 \times 10^{-34}\,\text{Js}\), \(e = 3.0 \times 10^{8}\,\text{m/s}\), I eV = \(1.6 \times 10^{-19}\,\text{J}\)]
(a)(i) Principal factor for nuclear stability. The relative stability of a nucleus is determined principally by its neutron-to-proton (n/p) ratio (equivalently, by its binding energy per nucleon). Nuclei whose n/p ratio lies within the stability band are stable; those far from it are unstable.
(a)(ii) Order of stability. Compare the neutron-to-proton ratios of the given nuclides. A nucleus whose n/p ratio is closest to 1 (for light nuclei) or lies nearest the stability line is the most stable, and stability decreases as the n/p ratio departs further from it. Arrange the nuclides so that the one with the n/p ratio nearest the stability band comes first and the one furthest from it comes last (for example the light nuclide with n/p close to 1 is the most stable, and the heavy neutron-rich nuclide is the least stable). Justify each placement by quoting its computed n/p ratio.
(b)(i) Ionization potential. The ionization potential is the minimum energy (or the potential difference through which an electron must be accelerated to acquire that energy) required to completely remove the most loosely bound electron from an isolated atom in its ground state.
(b)(ii) Ground-state energy \(E_0\). The photon emitted in the transition \(n=3 \to n=1\) carries energy \(hc/\lambda\). (Here \(c = 3.0\times10^{8}\ \text{m/s}\).)
\[ E_{3}-E_{1} = \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\times3.0\times10^{8}}{1.02\times10^{-7}} = 1.94\times10^{-18}\ \text{J} \]
For hydrogen \(E_n = \dfrac{E_0}{n^{2}}\) with \(E_0\) the ground-state energy, so the emitted energy is \(|E_0|\left(1-\tfrac{1}{9}\right)=\tfrac{8}{9}|E_0|\):
\[ \tfrac{8}{9}|E_0| = 1.94\times10^{-18} \;\Rightarrow\; |E_0| = 2.18\times10^{-18}\ \text{J} \]
So \(E_0 \approx -2.18\times10^{-18}\ \text{J}\) (about \(-13.6\ \text{eV}\)).
Ionization potential of hydrogen. The energy to remove the electron from the ground state is \(|E_0| = 2.18\times10^{-18}\ \text{J}\); in volts,
\[ V = \frac{|E_0|}{e} = \frac{2.18\times10^{-18}}{1.6\times10^{-19}} \approx 13.6\ \text{V} \]
(c)(i) Work function of sodium is 2.0 eV. This means the minimum energy needed to just free an electron from the surface of sodium metal is 2.0 electron-volts \((= 2.0\times1.6\times10^{-19} = 3.2\times10^{-19}\ \text{J})\).
(c)(ii) Photoelectron emission (\(\lambda = 160\ \text{nm} = 1.6\times10^{-7}\ \text{m}\)). Energy of the incident photon:
\[ E = \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\times3.0\times10^{8}}{1.6\times10^{-7}} = 1.24\times10^{-18}\ \text{J} = 7.7\ \text{eV} \]
Since the photon energy (7.7 eV) is greater than the work function (2.0 eV), photoelectrons will be emitted, each with maximum kinetic energy \(7.7 - 2.0 = 5.7\ \text{eV}\).
(a)(i) Principal factor for nuclear stability. The relative stability of a nucleus is determined principally by its neutron-to-proton (n/p) ratio (equivalently, by its binding energy per nucleon). Nuclei whose n/p ratio lies within the stability band are stable; those far from it are unstable.
(a)(ii) Order of stability. Compare the neutron-to-proton ratios of the given nuclides. A nucleus whose n/p ratio is closest to 1 (for light nuclei) or lies nearest the stability line is the most stable, and stability decreases as the n/p ratio departs further from it. Arrange the nuclides so that the one with the n/p ratio nearest the stability band comes first and the one furthest from it comes last (for example the light nuclide with n/p close to 1 is the most stable, and the heavy neutron-rich nuclide is the least stable). Justify each placement by quoting its computed n/p ratio.
(b)(i) Ionization potential. The ionization potential is the minimum energy (or the potential difference through which an electron must be accelerated to acquire that energy) required to completely remove the most loosely bound electron from an isolated atom in its ground state.
(b)(ii) Ground-state energy \(E_0\). The photon emitted in the transition \(n=3 \to n=1\) carries energy \(hc/\lambda\). (Here \(c = 3.0\times10^{8}\ \text{m/s}\).)
\[ E_{3}-E_{1} = \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\times3.0\times10^{8}}{1.02\times10^{-7}} = 1.94\times10^{-18}\ \text{J} \]
For hydrogen \(E_n = \dfrac{E_0}{n^{2}}\) with \(E_0\) the ground-state energy, so the emitted energy is \(|E_0|\left(1-\tfrac{1}{9}\right)=\tfrac{8}{9}|E_0|\):
\[ \tfrac{8}{9}|E_0| = 1.94\times10^{-18} \;\Rightarrow\; |E_0| = 2.18\times10^{-18}\ \text{J} \]
So \(E_0 \approx -2.18\times10^{-18}\ \text{J}\) (about \(-13.6\ \text{eV}\)).
Ionization potential of hydrogen. The energy to remove the electron from the ground state is \(|E_0| = 2.18\times10^{-18}\ \text{J}\); in volts,
\[ V = \frac{|E_0|}{e} = \frac{2.18\times10^{-18}}{1.6\times10^{-19}} \approx 13.6\ \text{V} \]
(c)(i) Work function of sodium is 2.0 eV. This means the minimum energy needed to just free an electron from the surface of sodium metal is 2.0 electron-volts \((= 2.0\times1.6\times10^{-19} = 3.2\times10^{-19}\ \text{J})\).
(c)(ii) Photoelectron emission (\(\lambda = 160\ \text{nm} = 1.6\times10^{-7}\ \text{m}\)). Energy of the incident photon:
\[ E = \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\times3.0\times10^{8}}{1.6\times10^{-7}} = 1.24\times10^{-18}\ \text{J} = 7.7\ \text{eV} \]
Since the photon energy (7.7 eV) is greater than the work function (2.0 eV), photoelectrons will be emitted, each with maximum kinetic energy \(7.7 - 2.0 = 5.7\ \text{eV}\).
