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

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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^-13J

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^-13J

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 × t_1/2 = 0.693

Decay constant λ= 0.693 / t_1/2

The decay constant needs to be given in s^-1. So, the half-life should be in second.

Half-life t_1/2= (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 × 10²³particles.

So, mass of 6.02 × 10²³particles = 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 = P₀exp (-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