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.