Question

WORD PROBLEM - Consider the alpha decay of uranium-238,

238 92 U ?234 90 T h + 4 2 He

What is the value of the binding energy per nucleon of uranium-238 (in MeV)?

What is the value of the disintegration energy, i.e., the energy released in this decay (in MeV)?

Given: the atomic mass of uranium-238 is 238.050784 u, of thorium-234 is 234.036596 u, of helium-4 is 4.002603 u.

Answer 1

atomic mass of uranium-238 is = 238.050784 a.m.u;

thorium-234 is = 234.036596 a.m.u;

helium-4 is = 4.002603 a.m.u;

proton is =1.0072766 a.m.u;

neutron is = 1.0086654 a.m.u.

a) The sum of the masses of the 92 protons and 146 neutrons in uranium-238 is;

(92)* (1.0072766 a.m.u.) + (146)* (1.0086654 a.m.u.) = 239.9345956 a.m.u;

This is in excess of the atomic mass of uranium-235. The total nuclear binding energy is
thus,

dm = 239.9345956 a.m.u – 238.050784 a.m.u = 1.8838116 a.m.u;

Binding energy (Eb) = dm*(931.494) MeV = 1754.759203 MeV.

The binding energy per nucleon is therefore 1754.759203/ 238 = 7.3729 MeV/nucleon.

(b)  energy released in this decay

let m_i = initial mass, m_f = final mass

m_i - m_f = (238.050784 - (234.036596 + 4.002603)) a.m.u = 0.011585 a.m.u

energy relesed (Q) = (0.011585)*(931.494) = 10.79 MeV.