Nuclear fusion energy released

http://filestore.aqa.org.uk/subjects/AQA-PHYA5-1-QP-JUN13.PDF

Could someone pls help with question 1 c ii, I have no idea how they get the answer of 24.7 MeV

Firstly, you need to find the values of H, He and e+ in terms of 'u'. He is already given in the question, use the AQA data sheet to find the value of e+ in terms of u (It is stated under the electron rest mass). You also need the 'u' equivalent for H, because the hydrogen nucleus only has a proton in it, you can use the 'u' equivalent of a proton, this can be found on the AQA data sheet too.

When all these values have been found they can replace the elements in the reaction to make a simple equation. Be aware that the 2 electron neutrinos (Ve) are massless, so no 'u' equivalent is needed.

(4 x 'u' of a proton) - 4.00150u - (2 x 'u' of an electron)

this gives

(4 x 1.00728)u - 4.00150u - (2 x 5.5x10^-4)u = 0.02652u

Using the conversion that 1u=931.5MeV
0.02652 x 931.5 = 24.7MeV

Hope this helped!