Q#20 (Past Exam Paper – June 2010 Paper 41 Q4)
(a) Explain what is meant by the potential energy of a body. [2]
(b) Two deuterium (21H) nuclei each have initial kinetic energy EK and are initially separated by a large distance.
The nuclei may be considered to be spheres of diameter 3.8 × 10-15 m with their masses
and charges concentrated at their centres.
The nuclei move from their initial positions to their final position of just touching, as
illustrated in Fig. 4.1.
Fig. 4.1
(i) For the two nuclei approaching each other, calculate the total change in
1. gravitational potential energy, [3]
2. electric potential energy. [3]
(ii) Use your answers in (i) to show that the initial kinetic energy EK of each nucleus
is 0.19 MeV. [2]
(iii) The two nuclei may rebound from each other. Suggest one other effect that could
happen to the two nuclei if the initial kinetic energy of each nucleus is greater than
that calculated in (ii). [1]
Solution:
(a) The potential energy of a body is the ability of the body to do work as a result of its position or shape.
(b)
(i)
1.
ΔEgpe = GMm / r
{Deuterium consists of a proton and a neutron, each having a mass of 1.66×10-27 kg.
So, the mass of deuterium is (2 × 1.66×10-27) kg.}
ΔEgpe = (6.67×10-11 × {2 × 1.66×10-27}2) / (3.8×10-15)
ΔEgpe = 1.93×10-49 J
2.
ΔEepe = Qq / 4πε0 r
{Each deuterium consists of a proton and thus, have a charge of 1.6×10-19 C.}
ΔEepe = (1.6×10-19)2 / (4π × 8.85×10-12 × 3.8×10-15)
ΔEepe = 6.06×10-14 J
(ii)
{Initially, each nucleus have kinetic energy EK and GPE.
Total initial kinetic energy = EK + EK + GPE = 2EK + GPE
As the two nuclei approach each other, their energies are converted into EPE. So,
2EK + ΔEgpe = ΔEepe
2EK = ΔEepe – ΔEgpe
The change in GPE is very much smaller compared to the change in EPE. So,
ΔEepe – ΔEgpe = 6.06×10-14 J
2EK = 6.06×10-14 J
EK = 6.06×10-14 / 2 = 3.03×10-14 J
To convert J to MeV, we divide by 1.6×10-13.
EK = 3.03×10-14 / 1.6×10-13
EK = 0.19 MeV}
(iii) fusion may occur / may break into sub-nuclear particles
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