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Question

Two protons and two neutrons combine to form a nucleus of $$_{2}{He}^{4}$$. Find the energy released during the process. Take masses of proton, neutron and helium nuclues as $$1.007  u,  1.009  u,  4.002  u$$ respectively.


A
27.94 MeV
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B
84.67 MeV
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C
94.82 MeV
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D
24.12 MeV
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Solution

The correct option is A 27.94 MeV
The nuclear reaction is $$2 _{1}{p}^{1} + 2 _{0}{n}^{1}  \rightarrow _{2}{He}^{4}$$
First, let us find mass defect and then energy.
The mass of $$2p$$ and $$2n$$ is $$\left(2 \times 1.007 + 2 \times 1.009\right) = 2 \times 2.016 = 4.032  u$$.
The mass of helium is $$= 4.002  u$$.
The mass defect $$= 4.032 - 4.002 = 0.030  u$$.
$$1  u$$ liberates $$931.5  MeV$$ of energy. The energy equivalent to $$0.030  u = 0.03 \times 931.5 = 27.94  MeV$$.
The above nuclear reaction is called fusion as lighter nuclei combine together to form a single nuclei.

Physics

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