Question

When a lithium-7 nucleus is bombarded by a proton, two alpha particles are produced. The Q value is
(1H1: 1.007825 u, 2He4: 4.002603 u, 3Li7: 7.016004 u)​

Answer 1

Given:

1H1=1.008

2He4=4.003

3Li7=7.016

To find:

The Q- value

Solution:

The formula of Q is

Q= Δm×931 MeV

Q=[m(Li)+m(H)-2m(He)]×931

Q=[7.016+1.008-2×4.003]×931

Q=0.016×931

Q=14.896 MeV

So, the value of Q is 14.896 MeV

hope it helps :)

Answer 2

Explanation:

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