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Problem 9: Find the energy released in when β-decay changes 90Th 234 into 91Pa234. Mass of 90Th 234 = 234.0436 u and 91Pa234 = 234.042762 u.

Theory

Equation of the nuclear reaction is as follow;

90Th 234 —–→ 91Pa234 + β0 + anti-neutrino + Q (energy)

Physically, the unstable 90Th 234 nucleus emits an electron and an anti-neutrino and converts to 91Pa234 with the release of some energy. At the same time, a neutron in the nucleus converts to proton. Thus there is no change in the mass number A. However, the charge number increases by 1 unit.

Solution

To find the energy released, we have to find the mass defect.

Now,     Mass of 90Th 234 = 234.0436 u = Mass on LHS of the equation.

Mass of 91 Pa 234 = 234.042762 u

Mass of β0 = 0.000547 u

Therefore, mass on RHS = 234.042762 u + 0.000547 u = 234.043309 u

Mass defect, Δ m = 234.0436 u – 234.043309 u = 0.000291 u

Now we know that 1 u = 931.5 MeV,

⇒ 0.000291 × 931.5 = 0.2710665 MeV

3 Comments

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