Question

The binding energy of a heavy nucleus is about $7$ million electron volts per nucleon, whereas the binding energy of a medium-weight nucleus is about $8$ million electron volts per nucleon. Therefore, the total kinetic energy liberated when a heavy nucleus undergoes symmetric fission is most nearly:

• A. $1876MeV$
• B. $938MeV$
• C. $200MeV$
• D. $8MeV$
• E. $7MeV$

Answer 1

The correct option is C $200MeV$

Symmetric fusion reaction can be written as below:

${}^{A}X\to 2{}^{A/2}Y$

Change in binding energy per nucleon equals the kinetic energy liberated.

$\Delta E=2\frac{E_f}{2}-E_i=E_f-E_i$

where $E_f$ & $E_i$ are binding energies of $Y$ and $X$ respectively.

A heavy nucleus has approximately $200$ nucleons.

Hence, total energy liberated is

$E\approx 200\Delta E=200\times (8\times {10}^6-7\times {10}^6)$

$E=200MeV$