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Feliz [49]
3 years ago
7

How much energy is released when a proton combines with a deuterium nucleus to produce 3He?

Physics
1 answer:
slava [35]3 years ago
6 0

Answer:

E =  7.99 *10^{-13} J

Explanation:

the given reaction is

_1 H^1 + _1 H^2 = _2 He^3

we know that energy is given asE = \Delta mc^2E = (m_1 H^1 + m_1 H^2  - _2 He^3)c^2

where

m_1 H^1 is mass of proton = 1.672622 *10^{-27}

m_1 H^2 is mass of deuterium = 3.344494 *10^{-27}

m_2 H^3 is mass of He = 5.008234 *10^{-27}

E = [1.672622 *10^{-27} + 3.344494 *10^{-27} - 5.008234 *10^{-27} ] *(3*10^8)^2

E =  7.99 *10^{-13} J

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Cheers!

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