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

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