Q.The fission properties of $^{239}_{94}\text{Pu}$ are very similar to those of $^{235}_{92}\text{U}$. The average energy released per fission is $180\ \text{MeV}$. How much energy, in MeV, is released if all the atoms in $1\ \text{kg}$ of pure $^{239}_{94}\text{Pu}$ undergo fission?

Concept: Mass–Energy Equivalence – the energy released is the number of fissions multiplied by the energy per fission.

Step 1: Number of atoms in 1 kg of $^{239}_{94}\text{Pu}$

Molar mass $M = 239\ \text{g/mol} = 0.239\ \text{kg/mol}$.

Number of moles $n = \frac{1\ \text{kg}}{0.239\ \text{kg/mol}} \approx 4.184\ \text{mol}$.

Avogadro’s number $N_A = 6.022 \times 10^{23}\ \text{atoms/mol}$.

Total atoms $N = n \times N_A \approx 4.184 \times 6.022 \times 10^{23} \approx 2.52 \times 10^{24}$.

Step 2: Total energy released …

The core idea here is mass–energy equivalence — but not in the way you might first think. We aren’t directly converting the entire mass of plutonium into energy (that would be annihilation, not fission). Instead, each fission event converts a tiny fraction of the nucleus’s mass into kinetic energy of fragments and neutrons, which we measure as 180 MeV per fission. To find the total energy from 1 kg, we simply count how many fission events happen and multiply.

Let’s walk through it step by step.

1. Find the number of atoms in 1 kg of $^{239}\text{Pu}$. The molar mass of $^{239}\text{Pu}$ is approximately $239\ \text{g/mol}$ (since the atomic mass number is 239). One mole contains Avogadro’s number of atoms, $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$. For 1 kg = 1000 g, the number of moles is:

So the number of atoms is:

2. Multiply by the energy per fission. Each fission releases 180 MeV. Therefore, total energy:

Method: Direct calculation using Avogadro’s number and mass–energy equivalence.

The idea is simple: find the number of atoms in 1 kg of plutonium-239, then multiply by the energy released per fission.

Step 1 – Find the number of moles in 1 kg of $^{239}\text{Pu}$.

The molar mass of $^{239}\text{Pu}$ is $239\ \text{g/mol}$ (since the mass number is 239).

$1\ \text{kg} = 1000\ \text{g}$, so

Step 2 – Find the number of atoms.

Avogadro’s number $N_A = 6.022 \times 10^{23}$ atoms/mol.

Step 3 – Multiply by the energy per fission.

Each fission releases $180\ \text{MeV}$.

Step 4 – Calculate.

First, $\frac{1000}{239} \approx 4.1841$. …

Common Mistakes & How to Avoid Them

Mistake 1: Using the wrong mass number or atomic mass

Students often take the mass number (239) as the exact atomic mass in grams, or they confuse it with the molar mass in g/mol. The mass number is approximately the molar mass, but the real atomic mass of $^{239}\text{Pu}$ is 239.05216 u — close, but not exactly 239.

How to avoid: For exam problems, when the exact atomic mass is not given, use the mass number (239) as the molar mass in g/mol. This is the standard approximation in such questions. So 1 mole of $^{239}\text{Pu}$ has a mass of 239 g.

Mistake 2: Forgetting to convert kg to g

The problem gives mass in kg (1 kg), but the molar mass is in g/mol. Students sometimes plug 1 kg directly into the formula without converting.

How to avoid: Always check units. Convert 1 kg = 1000 g before using the molar mass.

Mistake 3: Confusing number of atoms with number of moles

Some students calculate the number of moles correctly but then forget to multiply by Avogadro's number to get the number of atoms.

How to avoid: Remember the chain:

Mistake 4: Using the wrong value of Avogadro's number

Using $6.022 \times 10^{23}$ is fine, but some students use $6.023 \times 10^{26}$ (which is for kg-mole) or forget the exponent entirely.

How to avoid: Stick to $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ for gram-mole calculations.

Mistake 5: Arithmetic errors in the final multiplication

The numbers are large — $1000/239 \approx 4.184$, multiplied by $6.022 \times 10^{23}$, then by 180. Students often misplace decimal points or exponents.

How to avoid: Do the calculation step by step, and keep track of powers of 10 separately.

Correct Solution …