E = mc2 - mass energy conversion
Q4.
(a)
(i) Copy and complete the equation below to represent the emission of an α particle by a 
isotope.
(ii) Calculate the energy released when a 
isotope nucleus emits an alpha particle
mass of uranium 238 nucleus: 238.05076 u
mass of an alpha particle: 4.00260 u
mass of thorium 234 nucleus: 234.04357 u
Δm = 238.05076 – (4.00260 + 234.04357) = 0.00459(u) 
Q = 931.3 × 0.00459 (MeV)
Q = 4.3MeV
(5 marks)
(b) decays sequentially by emitting α particles and β– particles, eventually forming 
, a stable isotope of lead.
(i) There are eight alpha particles in the sequence. Calculate the number of beta- particles in the sequence.
The overall change in proton number = 92 – 82 = 10
The change in proton number due to alpha particles = 8 × 2 = 16
Therefore ΔZ = –6 for the beta particles corresponding to the six beta particles
(ii) State the nuclear change that occurs during positron emission. Hence, explain why no positrons are emitted in this sequence.
In positron emission a proton changes to a neutron plus a positron and a neutrino.
This will increase the neutron to proton ratio of a nucleus.Alpha emission involves a loss of two protons and two neutrons but, as the number of neutrons in heavier nuclei is greater than the number of protons, loss of an alpha particle will also raise the neutron to proton ratio, but less than positron emission. Positron emission competes with alpha emission when the sequence reqires an increase in the ratiobut is energetically less favourable as overall a decrease in the ratio os required
(6 marks Max)
(Total 11 marks)
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