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Wednesday, March 4, 2020

The power for a space probe is to be supplied by the energy released when plutonium-236 decays by the emission of α-particles.


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}

t = 3.3 years

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