For ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n, use masses 235.0439 u, 1.0087 u, 140.9144 u and 91.9262 u respectively. Find energy released, taking 1 u c² = 931.5 MeV.

Solution & explanation

1. Initial mass includes uranium and the incoming neutron.

   $$m_i=1.0087+235.0439=236.0526$$

2. There are three outgoing neutrons.

   $$m_{\mathrm{n3}}=3\times 1.0087=3.0261$$

3. Add both fragments and all outgoing neutrons.

   $$m_f=3.0261+91.9262+140.9144=235.8667$$

4. Mass defect is initial minus final mass.

   $${\Delta }_m=236.0526-235.8667=0.1859$$

5. Convert mass defect in u directly to energy in MeV.

   $$E=0.1859\times 931.5$$

6. Multiply and round.

   $$E=173.16585$$

E ≈ 173.17 MeV with the supplied rounded masses