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Problem 8: 3Li6 is bombarded by deuteron. The reaction gives two α particles along with the release of energy equal to 22.3 MeV. Knowing masses of deuteron and α particles, determine mass of lithium isotope of 3Li6.

Theory

The energy released in the nuclear reaction is equal to the difference in masses of the reactants and products. As the quantity of the energy released in the reaction is given, we can calculate the isotope (atomic mass of the Lithium used).

Solution

The balanced reaction is

1H2 + 3Li6 —–→ 2(2He4) + 22.3 MeV

Now mass of deuterium = 2.014102 u

Mass of two helium atoms = 2(4.00388) = 8.00776 u

Similarly, 931.5 MeV = 1 u

⇒ 1 MeV = (1/931.5) u ⇒ 22.3 MeV = (1/931.5) × 22.3 u

Or 23.3 MeV = 0.0239399 u

So equating mass on RHS and LHS of the equation,

2.014102 + 3Li6 = 8.00776 + 0.0239399

3Li6 = (8.00776 + 0.0239399) – 2.014102

3Li6 = 6.017 u

3 Comments

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