Question

228Th emits an alpha particle to reduce to 224Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay
Th228→Ra224*+αRa224*→224Ra+Υ(217 keV).
Atomic mass of 228Th is 228.028726 u, that of 224Ra is 224.020196 u and that of H42 is 4.00260 u.

Answer 1

Explanation:

In the question, it is given:

228Th has the atomic mass, m(228Th) = 228.028726 u

224Ra has the atomic mass, m(224Ra) = 224.020196 u

H24 has the atomic mass, m(H24) = 4.00260 u

224Ra has the mass = $931 \times 224.020196+0.217 \mathrm{MeV}=208563.0195 \mathrm{MeV}$

Alpha particle’s kinetic energy, K =  mTh228 - mRa224 + mH24c2

= $-[(+4.00260+208563.0195 \times 931]+(228.028726 \times 931)$

= 5.304 MeV