Question 17
The power for a space
probe is to be supplied by the energy released when plutonium-236 decays by the
emission of α-particles.
The α-particles,
each of energy 5.75 MeV, are captured and their energy is converted into
electrical energy with an efficiency of 24%.
(a)
Calculate
(i)
the energy, in joules, equal to 5.75 MeV, [1]
(ii)
the number of α-particles per second required to generate 1.9
kW of electrical power. [2]
(b)
Each plutonium-236 nucleus, on disintegration, produces one α-particle.
Plutonium-236 has a
half-life of 2.8 years.
(i)
Calculate the decay constant, in s–1, of plutonium-236. [2]
(ii)
Use your answers in (a)(ii) and (b)(i)
to determine the mass of plutonium-236 required for the generation of
1.9 kW of electrical power. [4]
(c)
The minimum electrical power required for the space probe is 0.84
kW.
Calculate the time, in
years, for which the sample of plutonium-236 in (b)(ii)
will provide
sufficient power. [2]
Reference: Past Exam Paper – June 2015 Paper 42 Q8
Solution:
(a)
(i)
{To convert
eV to J, multiply by 1.6×10-19
.
To convert
MeV to J, multiply by 1.6×10-19,
then by 106 (that is, multiply by 1.6×10-13).}
energy =
5.75 × 1.6×10-13
energy =
9.2 × 10-13 J
(ii)
{1
particle generates 9.2×10-13
J
Power =
energy / time
Energy =
Pt
In 1 s,
energy = 1900 × 1 = 1900
J.
So, the
total electrical energy generated is 1900 J.
We need to
take into account the efficiency of the process.
Efficiency
= 24 % = 0.24
0.24 - - -
> 1900 J
Total
energy from particles = 100% = 1
1 - - -
> 1900 / 0.24 (= total energy)
BUT one
particle generates 9.2×10-13
J
Number of
particles = total energy / energy of 1 particle}
number =
1900 / (9.2×10-13 × 0.24)
number =
8.6 × 1015 s-1
(b)
(i)
{λ × t1/2 = 0.693
Decay constant λ= 0.693 / t1/2
The decay
constant needs to be given in s-1. So, the half-life should be in second.
Half-life t1/2
= (2.8 × 365 × 24 × 3600) s}
decay
constant = 0.693 / (2.8 × 365 × 24 × 3600)
decay
constant = 7.85 × 10-9 s-1
(ii)
{To find
the mass, we need to obtain the initial number of nuclei first.}
A = λN
8.6×1015 = 7.85×10-9 × N
{Number of
nuclei: N = 8.6×1015
/ 7.85×10-9}
N = 1.096 × 1024
{Plutonium-236:
Mass of 1
mole of plutonium = 236 g
1 mole
contains 6.02 × 1023
particles.
So, mass
of 6.02 × 1023
particles = 236 g
Mass of 1
particle = 236 / 6.02 × 1023
Mass of 1.096
× 1024 particles = (1.096×1024 × 236) /
(6.02×1023)}
mass =
(1.096×1024 × 236) /
(6.02×1023)
mass = 430
g
(c)
{The
radioactive source of plutonium decreases exponentially with time. So, the
initial power and final power should be related by an exponential equation.
P = P0
exp (-λt)
Where P is
the power at time t and P0 is the total power available from the particles.}
0.84 = 1.9
exp(–7.85×10-19
× t)
t = 1.04 × 108 s
{To convert
- Years to
seconds: multiply by 365 × 24 × 3600
- Seconds
to years: divide by 365 × 24 × 3600}
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