Ncert Exercises & Solutions

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

Hint: The Q-value of a reaction is given by \(Q=\Delta mc^2\).
Step 1: Find the fission reaction of \({ }_{26}^{56} \mathrm{Fe}\).
The fission of

{}_{26}^{56}F e

can be given as:
\(_{26}^{56}\mathrm{Fe} \rightarrow 2{ }_{13}^{28} \mathrm{Al}\)

Step 2: Find the Q-value of the fission reaction.
It is given that:

Atomic mass of m

({}_{26}^{56}F e)

=55.9349 u

Atomic mass of m

({}_{13}^{28}A l)

=27.98191 u

The Q-value of this nuclear reaction is gien as:

Q = [m ({}_{26}^{56}F e) - 2 m ({}_{13}^{28}A l)] c^{2}

= [55.93494 - 2 \times 27.9819] c^{2}

= - (- 0.02886 c^{2}) u

B u t 1 u = 931.5 M e V / c^{2}

∴ Q = - 0.0288 \times 931.5 = - 26 .8272 M eV

The Q-value of the fission is negative. Therefore, fission is not possible energetically. For an energetically-possible fission reaction, the Q-value must be positive.

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