Q.How long can an electric lamp of $100\ \text{W}$ be kept glowing by fusion of $2.0\ \text{kg}$ of deuterium? Take the fusion reaction as
$^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \rightarrow {}^{3}_{2}\text{He} + n + 3.27\ \text{MeV}$.

Concept: Mass–Energy Equivalence — fusion energy comes from the mass defect, released as $3.27\ \text{MeV}$ per D–D reaction.

Step 1: Energy per reaction: $3.27\times1.602\times10^{-13} = 5.24\times10^{-13}\ \text{J}$.

Step 2: Atoms of deuterium in $2.0\ \text{kg}$: $\dfrac{2000}{2}\times6.022\times10^{23} = 6.022\times10^{26}$; each reaction uses 2 atoms, so reactions $=3.011\times10^{26}$.

Step 3: Total energy: $3.011\times10^{26}\times5.24\times10^{-13} = 1.577\times10^{14}\ \text{J}$. …

Why this approach works

Deuterium nuclei fuse in pairs, each pair releasing $3.27\ \text{MeV}$. The lamp consumes $100\ \text{J}$ every second. So the question reduces to: how many fusion events can $2.0\ \text{kg}$ of deuterium supply, and how long will that energy last at $100\ \text{W}$? The only subtlety: each fusion event consumes two deuterium nuclei, so the number of reactions is half the number of atoms.

Step-by-step solution

1. Number of deuterium atoms in $2.0\ \text{kg}$

Deuterium ($^2_1\text{H}$) has a molar mass of about $2.0\ \text{g/mol}$ (mass number 2). So:

2. Number of fusion reactions

Each reaction uses two deuterium nuclei:

3. Total energy released

Each reaction releases $3.27\ \text{MeV}$; using $1\ \text{MeV}=1.602\times10^{-13}\ \text{J}$:

$E_{\text{total}} = (3.011\times10^{26})\times(5.238\times10^{-13}) = 1.577\times10^{14}\ \text{J}$ …

Method: Energy conservation via mass-energy equivalence (Einstein’s relation $E = \Delta m c^2$).

Step 1 – Find the number of deuterium nuclei in 2.0 kg.

Deuterium ($^2_1\text{H}$) has a molar mass of approximately $2.0\ \text{g/mol}$.

Number of moles in $2.0\ \text{kg}$:

Number of nuclei:

Step 2 – Determine the energy released per fusion event.

Each fusion reaction involves two deuterium nuclei and releases $3.27\ \text{MeV}$.

So the number of fusion events possible from $N$ nuclei is $N/2$ (since two nuclei are consumed per event).

Step 3 – Calculate total energy released.

Total energy in MeV:

Convert to joules ($1\ \text{MeV} = 1.602 \times 10^{-13}\ \text{J}$):

Step 4 – Relate energy to lamp power. …

Common Mistakes & How to Avoid Them

1. Forgetting to convert MeV to joules

The energy released per fusion event is given in MeV, but power is in watts (J/s). Many students plug 3.27 MeV directly into the calculation without converting to joules.

How to avoid: Always write the conversion factor explicitly before starting:

$1\ \text{MeV} = 1.6 \times 10^{-13}\ \text{J}$.

So $3.27\ \text{MeV} = 3.27 \times 1.6 \times 10^{-13}\ \text{J}$.

2. Using the wrong number of nuclei in 2 kg of deuterium

Deuterium is $^{2}_{1}\text{H}$, so its molar mass is 2 g/mol, not 1 g/mol. A common error is to treat it as ordinary hydrogen.

How to avoid:

Number of moles = $\frac{\text{mass}}{\text{molar mass}} = \frac{2000\ \text{g}}{2\ \text{g/mol}} = 1000\ \text{mol}$.

Number of nuclei = $1000 \times N_A = 1000 \times 6.02 \times 10^{23} = 6.02 \times 10^{26}$.

3. Forgetting that each reaction uses two deuterium atoms

Students sometimes take the total number of deuterium nuclei as the number of reactions. But the reaction $^{2}_{1}\text{H} + {}^{2}_{1}\text{H}$ needs two per event.

How to avoid:

Number of reactions = $\frac{\text{total deuterium nuclei}}{2} = \frac{6.02 \times 10^{26}}{2} = 3.01 \times 10^{26}$.

4. Confusing power with energy

Power is energy per second. The lamp consumes 100 J every second. The total energy from fusion gives the total time directly as $t = \frac{E_{\text{total}}}{P}$.

How to avoid:

Write the relation clearly:

$P = \frac{E}{t} \quad\Rightarrow\quad t = \frac{E}{P}$.

Do not multiply power and time — that gives energy, not time.

5. Arithmetic errors in powers of ten …