Question 371: Atoms & Nuclei

Question

A common example of alpha decay is $${ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2} \mathrm{He}^{4}+\mathrm{Q}$$
Given :
$${ }_{92}^{238} \mathrm{U}=238.05060 ~\mathrm{u}$$,
$${ }_{90}^{234} \mathrm{Th}=234.04360 ~\mathrm{u}$$,
$${ }_{2}^{4} \mathrm{He}=4.00260 ~\mathrm{u}$$ and
$$1 \mathrm{u}=931.5 \frac{\mathrm{MeV}}{c^{2}}$$
The energy released $$(Q)$$ during the alpha decay of $${ }_{92}^{238} \mathrm{U}$$ is __________ MeV

Accepted Answer

To find the energy released during the alpha decay, we first need to calculate the mass difference between the reactants and the products.
Mass difference = Mass of Uranium-238 - (Mass of Thorium-234 + Mass of Helium-4)
Mass difference = $$238.05060 \mathrm{u} - (234.04360 \mathrm{u} + 4.00260 \mathrm{u})$$
Mass difference = $$238.05060 \mathrm{u} - 238.04620 \mathrm{u}$$
Mass difference = $$0.00440 \mathrm{u}$$
Now, we can convert this mass difference to energy using the given conversion factor:
Energy released (Q) = Mass difference ÃÂ $$\frac{931.5 \mathrm{MeV}}{c^2}$$
Q = $$0.00440 \mathrm{u} ÃÂ 931.5 \frac{\mathrm{MeV}}{c^2}$$
Q Ã¢ÂÂ 4.1 MeV
The energy released (Q) during the alpha decay of $$^{238}U$$ is approximately 4.1 MeV.

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