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

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Solution

Q- value is given by,

Q = [m ({}_{26}^{56}{Fe}) - 2 m ({}_{13}^{28}{Al}) \times 931.5 MeV = [55.93494 - 2 \times 27.98191] \times 931.5 MeV

Q = - 0.02886 \times 931.5 MeV = - 26.88 MeV,

which is negative.

Since, the Q-value is negative (energy released from the reaction is negative), the fission is not possible energetically.

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