Question 20
(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]
Reference: Past Exam Paper – June 2010 Paper 41 Q4
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 / (4Ï€ × 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}
How do you know that initially the two nucleus have GPE ! It is not mentioned in question !
ReplyDeleteany object that has mass would exert a gravitational force. they also have some GPE in a gravitational field.
Deletewe have asked to calculate the GPE in (b)(i)1.
But why haven't we taken the square of radius for both epe n gpe
Deleteas given in the solution above, the formulae for GPE and EPE do not contain r^2. but only r
Delete