Find the kinetic energy of the α - particle emitted in the decay (in MeV) 238Pu→234U+α. The atomic masses needed are as follows: 238Pu=238.04955u 234U=234.04095u 4He=4.002603uNeglect residual nucleus

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Find the kinetic energy of the $α$ - particle emitted in the decay (in MeV)
$^{238} Pu \to^{234} U + α$ .
The atomic masses needed are as follows:
$\begin{array}{l} ^{238} Pu = 238 .04955 u \\ ^{234} U = 234 .04095 u \\ ^{4} He = 4 .002603 u \end{array}$
Neglect residual nucleus

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answer is 5.59.

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Detailed Solution

The mass defect in the process $Δm = m (^{238} Pu) - m (^{234} U) - m (^{4} He)$

$= 238 .04955 u - 234 .04095 u - 4 .002603 u = 0 .005997 u$

The kinetic energy of $α -$ particle $K = {Δmc}^{2} = 0 .005997 \times 931.5 MeV / u = 5 .586 MeV$

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