The NASA spacecraft Cassini relies on spontaneous decay of Plutonium 238 as a po

Question

The NASA spacecraft Cassini relies on spontaneous decay of Plutonium 238 as a power source (in a device called a Radioisotope Thermoelectric Generator). 238 Pu decays via alpha decay with a half-life of 87.7 years. Each decay event releases 5.70 MeV of energy. This energy is converted to electricity with an efficiency of 6.70%. The spacecraft requires a minimum of 500 W of electrical power. Find the minimum number of 238 Pu nuclei required for the spacecraft to operate.

Additional data. You may not need all of it.
1 eV = 1.602 x 10^-19 J
1 MeV = 106 eV
1 u (atomic mass unit) = 1.6605 x 10^-27 kg
Mass - energy equivalence: 1 u = 931.494 MeV
mp= mass of proton = 1.007825 u
mn = mass of neutron = 1.008665 u
me = mass of electron = 5.4858 x 10^-4
u
c = speed of light = 2.998 x 10^8 m s-1
? = Stefan-Boltzmann constant = 5.672 x 10^-8 J s-1 m-2 K-4
? = Earth's average planetary albedo = 0.34
rE = radius of planet Earth = 6370 km
Solar constant for Earth: S ? 1370 Wm-2
Nuclear masses :
U-235 235.04392 u
Barium-141 140.91441 u

Explanation / Answer

We know 500 W power = 500 joules per second of energy (in this case electrical energy)

Plutonium 238 decays via alpha emission in the NASA spacecraft Cassini in a device called a Radioisotope Thermoelectric Generator.

238Pu ------> 234?U + alpha + 5.7 MeV energy

Now each particle gives 5.7 MeV or 5.7 * 1.6 *10-19*106 = 9.12*10-13 J
Let X particles be required to provide 500 J/sec
Each particle provides 9.12*10-13 J of energy
So X * 9.12*10-13 J = 500
or X = 5.48 * 1014 particles

X constitutes 6.70% of total number of particles needed to be decayed per second (total efficiency)
So, grand total number of particles decayed X/6.70 * 100 = 14.92X = 76.58 * 1014 particles = A let.

Half life = 87.7 years = 2727820800 seconds = 2.7 * 109 seconds
Let we take initial concentration (i.e. number of particles) C.
C/2 remains after 2.7 * 109 seconds
C/(2*2.7 * 109) remains after 1 sec i.e. C/(2*2.7 * 109) decays in 1 sec.

Now as per situation demand:
C/(2*2.7 * 109) = 76.58 * 1014 particles = A
or C = 76.58 * 1014 * (2*2.7 * 109) = 347.67 * 1023 particles = 3.47 * 1025 particles.

So, minimum number of Pu needed = 3.47 * 1025 particles per second to provide required energy for running the spacecraft.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.