(ii) binding energy of a nucleus.
(b)(i) Assuming the wave nature of an electron, what is the effect of decreasing the speed of a photoelectron on its; (\(\alpha\)) wavelength? (\(\beta\)) energy?
(ii) A particle of friasS 4.4 x 10\(^{-23}\) kg moves with a velocity of 10\(^5\)ms\(^{-1}\). Calculate its wavelength. (h = 6.6 x 10\(^{-34}\) Js)
The diagram above shows part of a radioactive decay series. Use it to answer the following questions.
(i) Name a pair of isotopes.
(ii) Name the isotopes with which the series starts.
(iii) Write down a nuclear equation for two ekgmples of each of: (\(\alpha\)) alpha decay; (\(\beta\)) beta decay.
(a)(i) Mass defect. The mass defect is the difference between the sum of the separate masses of all the protons and neutrons (nucleons) in a nucleus and the actual (smaller) measured mass of the nucleus:
\[ \Delta m = \big(Z m_p + (A-Z) m_n\big) - M_{\text{nucleus}} \]
(a)(ii) Binding energy of a nucleus. It is the energy equivalent of the mass defect (\(E = \Delta m\, c^2\)); that is, the energy released when the free nucleons come together to form the nucleus, or equivalently the energy that must be supplied to separate the nucleus completely into its individual nucleons.
(b)(i) Effect of decreasing the speed of a photoelectron (de Broglie relation \(\lambda = h/mv\), kinetic energy \(E = \tfrac{1}{2}mv^2\)):
- (\(\alpha\)) Wavelength: since \(\lambda = \dfrac{h}{mv}\), decreasing \(v\) increases the wavelength.
- (\(\beta\)) Energy: since \(E = \tfrac{1}{2}mv^2\), decreasing \(v\) decreases the kinetic energy.
(b)(ii) de Broglie wavelength. Data: \(m = 4.4 \times 10^{-23}\ \text{kg}\), \(v = 10^{5}\ \text{m s}^{-1}\), \(h = 6.6 \times 10^{-34}\ \text{J s}\).
\[ \lambda = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{(4.4 \times 10^{-23})(10^{5})} \]\[ \lambda = \frac{6.6 \times 10^{-34}}{4.4 \times 10^{-18}} = 1.5 \times 10^{-16}\ \text{m} \]
(c) The decay-series graph. The graph plots nucleon number \(A\) (vertical, 208 to 232) against proton number \(Z\) (horizontal, Pb 82, Bi 83, Po 84, At 85, Rn 86, Fr 87, Ra 88, Ac 89, Th 90). Each down-left diagonal step (\(Z\) falls by 2, \(A\) falls by 4) is an alpha decay; each horizontal step to the right (\(Z\) rises by 1, \(A\) unchanged) is a beta decay. This is the thorium (4n) series running from Th-232 down to Pb-208.
(c)(i) A pair of isotopes. Isotopes have the same proton number but different nucleon numbers. From the graph, examples are the two lead points \(^{208}_{82}\text{Pb}\) and \(^{212}_{82}\text{Pb}\) (both at Z = 82); equally acceptable: \(^{216}_{84}\text{Po}\) and \(^{212}_{84}\text{Po}\) (Z = 84), or \(^{228}_{88}\text{Ra}\) and \(^{224}_{88}\text{Ra}\) (Z = 88), or \(^{232}_{90}\text{Th}\) and \(^{228}_{90}\text{Th}\) (Z = 90).
(c)(ii) Isotope with which the series starts. The top-right point, at Z = 90 and A = 232, is thorium-232, \(^{232}_{90}\text{Th}\).
(c)(iii) Nuclear equations (two examples of each).
(\(\alpha\)) Alpha decay (emits \(^{4}_{2}\text{He}\); Z falls by 2, A by 4):
\[ ^{232}_{90}\text{Th} \rightarrow\ ^{228}_{88}\text{Ra} + ^{4}_{2}\text{He} \]\[ ^{224}_{88}\text{Ra} \rightarrow\ ^{220}_{86}\text{Rn} + ^{4}_{2}\text{He} \]
(\(\beta\)) Beta decay (emits \(^{0}_{-1}e\); Z rises by 1, A unchanged):
\[ ^{228}_{88}\text{Ra} \rightarrow\ ^{228}_{89}\text{Ac} + ^{0}_{-1}e \]\[ ^{212}_{82}\text{Pb} \rightarrow\ ^{212}_{83}\text{Bi} + ^{0}_{-1}e \]
