Question

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

Answer 1

E = 7.99 *10^{-13} J

Step-by-step explanation:

the given reaction is

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

we know that energy is given as
`E = \Delta mc^2`
`E = (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