# Calculate the energy released in the nuclear reaction \\ce{_3^7Li + p ->2\alpha}\ given mass of \\ce{_3^7Li}\ atom and of helium atom to be 7.016 u and 4.0026 u respectively. - Physics

Numerical

Calculate the energy released in the nuclear reaction $\ce{_3^7Li + p ->2\alpha}$ given mass of $\ce{_3^7Li}$ atom and of helium atom to be 7.016 u and 4.0026 u respectively.

#### Solution

Data: M1 $\ce{(_3^7Li atom)}$ = 7.016 u, M2 = (He atom)

= 4.0026 u, mp = 1.00728 u, 1 u = 931.5 MeV/c2

Δ M = M1 + mp - 2M2

= (7.016 + 1.00728- 2(4.0026)]u

= 0.01808 u = (0.01808)(931.5) MeV/c2

= 16.84152 MeV/c2

Therefore, the energy released in the nuclear reaction= (ΔM)c2 = 16.84152 MeV

Concept: Nuclear Energy
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#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 15 Structure of Atoms and Nuclei
Exercises | Q 13 | Page 342