Calculate the amount of energy released during the α-decay of
Given:
atomic mass of ${}_{92}^{238}U = 238.05079 u$
atomic mass of ${}_{90}^{234}{Th} = 234.04363 u$
atomic mass of ${}_{2}^{4}{He} = 4.00260 u$
$1 u = 931.5 MeV / c^{2}$
${}_{92}^{238}U \to {}_{90}^{234}{Th} + {}_{2}^{4}{He}$ Is this decay spontaneous? Give reasons.

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Solution

The energy released in the α-decay is,

Q =

[m ({}_{92}^{238}U) - m ({}_{90}^{234}{Th}) - m ({}_{2}^{4}{He})] c^{2}

= [238.05079 - 234.04363 - 4.00260] \times 931.5 MeV = 0.00456 \times 931.5 = 4.25 MeV

Since Q-value is positive, the decay process is spontaneous.

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