Question

When a deuteron of mass 2.0141u and negligible kinetic energy is absorbed by lithium (⁶₃Li) of nucleus of mass 6.0155 u, the compound nucleus disintegrates spontaneously into two alpha particles, each of mass 4.0026 u.

Calculate the energy in joules carried by each alpha particle. (1 u = 1.66 x 10^-27 kg).

Answer 1

The reaction is:

²₁H + ⁶₃Li → ⁸₄Be → 2[⁴₂He]

∴ Mass defect, ∆m

= m (²₁H) + m (⁶₃Li) - 2m (⁴₂He)

= [2.0141 + 6.0155 - 2 x 4.0026] u

= 0.0244 u

 0.0244 x 1.66 x 10^-27 kg

= 0.0405 x 10^-27 kg

∴ Binding energy Q = ∆mc²

= 0.0405 x 10^-27 x 9 x 10¹⁶

= 0.3645 x 10^-11 J

= 36.45 x 10^-13 J

This energy is equally shared by both the alpha particles.

∴ Energy carried by each alpha particle

= \(\frac{1}{2}\) [36.45 x 10^-13] = 18.225 x 10^-13 J