Suppose, we think of fission of a

Question:

Suppose, we think of 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 $m\left({ }_{26}^{56} \mathrm{Fe}\right)=55.93494$ u and $m\left({ }_{13}^{28} \mathrm{Al}\right)=27.98191 \mathrm{u}$.

Solution:

The fission of ${ }_{26}^{56} \mathrm{Fe}$ can be given as:

${ }_{13}^{56} \mathrm{Fe} \longrightarrow 2{ }_{13}^{28} \mathrm{Al}$

It is given that:

Atomic mass of $m\left({ }_{26}^{56} \mathrm{Fe}\right)=55.93494 \mathrm{u}$

Atomic mass of $m\left({ }_{13}^{28} \mathrm{Al}\right)=27.98191 \mathrm{u}$

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

$Q=\left[m\left({ }_{26}^{56} \mathrm{Fe}\right)-2 m\left({ }_{13}^{28} \mathrm{Al}\right)\right] c^{2}$

$=[55.93494-2 \times 27.98191] c^{2}$

$=\left(-0.02888 c^{2}\right) \mathrm{u}$

But $1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}$

$\therefore Q=-0.02888 \times 931.5=-26.902 \mathrm{MeV}$

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.

Leave a comment

None
Free Study Material