Calculate the energy, in mega electron volts, released in...

Calculate the energy, in mega electron volts, released in the nuclear reaction $$_{5 \mathrm{B}}^{10} \mathrm{H}+_{2}^{4} \mathrm{He} \longrightarrow_{6}^{13} \mathrm{C}+_{1}^{1} \mathrm{H}$$ The nuclidic masses are $ \frac{10}{5}$ B=$10.01294 \mathrm{u}$.$_{2}^{4} \mathrm{He}=$ $4.00260 \mathrm{u}$;$_{2}^{4} \mathrm{He}=4.00260 \mathrm{u}$; $_{1}^{1} \mathrm{H}=1.00783 \mathrm{u}$.

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Key Concepts

Mass Defect

The mass defect is the difference between the total mass of the reactants and the total mass of the products in a nuclear reaction. This missing mass has been converted into energy, highlighting the direct relationship between mass and energy in nuclear processes.

Mass-Energy Equivalence

Mass-energy equivalence, encapsulated by Einstein’s famous equation E = mc², states that mass can be converted to energy. In nuclear reactions, even a small mass difference can result in a significant energy release, because the conversion factor (speed of light squared) is very large.

Atomic Mass Unit Conversion

The atomic mass unit (u) is a standard unit of mass used in nuclear physics, and it can be converted to energy units using the approximate conversion factor 1 u = 931.5 MeV. This conversion is essential in calculating the energy released in nuclear reactions.

Nuclear Reaction

A nuclear reaction involves changes in an atom’s nucleus, leading to a rearrangement of nucleons. Such reactions often result in the release or absorption of energy due to differences in nuclear binding energies, as evidenced by the mass defect and its conversion to energy.

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