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# During the fusion process, how is mass converted into energy?

Apr 15, 2016

$E = m {c}^{2}$

#### Explanation:

This is calculated using the famous equation of Einstein,

$E = m {c}^{2}$

In Fusion reaction like the ones taking place in the core of a Star, there is enough pressure to fuse hydrogen nuclei to form one helium nucleus.

So, 4 hydrogen nuclei are fused together to form one Helium nucleus. But, where does the energy come from that keeps the Sun from collapsing?.

When 4 Hydrogen nuclei are merged together they show a certain discrepancy in the mass when a Helium atom is formed, i.e the mass of 4 Hydrogen atoms before Fusion is higher than the mass of the Helium atom after the reaction this mass defect is converted into energy by $E = m {c}^{2}$.

$\text{mass of a hydrogen atom " = " 1.00794 u}$

$\text{mass of one helium atom " = " 4.002602 u}$

Where $u = 1.6605 \times {10}^{- 27} \text{kg}$.

The mass defect, $\Delta m$, will be

$\Delta m = 4 \times {m}_{H} - {m}_{H e}$

$= 4 \times \text{1.00794 u" - "4.002602 u}$

$= \text{0.029158 u }$ or $\text{ "4.8416859 * 10^(-29)"kg}$

In the equation $E = m {c}^{2}$, $c$ is the speed of light in a vacuum, approximately equal to $3 \cdot {10}^{8} {\text{m s}}^{- 1}$.

This means that you have

$E = 4.8416859 \cdot {10}^{- 29} {\text{kg" * (3 * 10^8)^2"m"^2"s}}^{- 2}$

$E = 4.35751731 \cdot {10}^{- 12} \text{ J per reaction}$