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  • A common example of alpha decay is <img alt="\begin{aligned} & { }_{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_2 \mathrm{He}^4+\mathrm{Q} \\ & \text { Given :

A common example of alpha decay is

\begin{aligned} & { }_{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_2 \mathrm{He}^4+\mathrm{Q} \\ & \text { Given : } \\ & { }_{92}^{238} \mathrm{U}=238.05060 \mathrm{u}, \\ & { }_{92}^{234} \mathrm{Th}=234.04360 \mathrm{u}, \\ & 90 \\ & { }_2^4 \mathrm{He}=4.00260 \mathrm{u} \text { and } \\ & 2 \\ & \text { lu }=931.5 \frac{\mathrm{MeV}}{\mathrm{c}^2} \end{aligned}

The energy released (Q) during the alpha decay of{ }_{92}^{238} \mathrm{U} is ________ Me V

Option: 1

4.25


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \alpha=\left(\mathrm{m}_{\mathrm{u}}-\mathrm{M}_{\mathrm{th}}-\mathrm{M}_{\mathrm{He}}\right) \mathrm{C}^2 \\ & \alpha=0.00456 \times 931.5 \\ & \alpha=4.25 \mathrm{MeV} \end{aligned}

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