Question

Suppose we think of fission of a \(^{56}_{26}Fe\) nucleus into two equal fragments, \(^{28}_{13} Al.\) Is the fission energetically possible? Argue by working out Q of the process. Given \(m\)(\(^{56}_{26}Fe\)) = 55.93494 u and \(m\)(\(^{28}_{13} Al\)) = 27.98191 u.

Answer 1

The fission of \( ^{56}_ {26}Fe \) can be given as: 
\(^{56}_{13} Fe → 2\space ^{28}_{13} Al\)
It is given that:
Atomic mass of \(m(^{56}_ {26}Fe) = 55.93494 u\)
Atomic mass of \(m ( ^{28}_{13} Al ) = 27.98191 u\)
The Q-value of this nuclear reaction is given as:
Q = \([m(^{56}_{26}Fe) - 2m(^{28}_{13} Al)]c^2\)
\(Q = [55.93494 - 2 \times  27.98191]c^2\)
\(Q = (-0.02888 \space c^2)u\)
But 1u = 931.5 \(\frac{MeV}{c^{2}}\)
Q = -0.02888 x 931.5
Q = -26.902 MeV 
The Q-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically possible fission reaction, the Q-value must be positive.