QJEEM23Phc815: JEE Main 2023 Physics - Class 12 JEEM23Phc815 Solution: A common example of alpha decay is 92 238 U ⟶ 90 234 Th + 2 He…

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 × $$\frac{931.5\mathrm{MeV}}{c^2}$$

Q = $$0.00440\mathrm{u}\times 931.5\frac{\mathrm{MeV}}{c^2}$$

Q ≈ 4.1 MeV

The energy released (Q) during the alpha decay of $${}^{238}U$$ is approximately 4.1 MeV.