4. The radius of a Uranium-238 nucleus is approximately 7.8 x 10-15 m.Using Heisenberg's uncertainty relation, estimate the energy of an a particlethat can be emitted from the nucleus, assuming that the a particle waslocalized inside the nucleus before it emerged. Use sensible units.

Given Data:Radius of U92238 nucleus (r) = 7.8×10-15 m. If the alpha particle is localized within the nucleus, this should mean that ∆x=2r =2×7.8×10-15 m= 1.56×10-14 m (It can be anywhere within the diameter of the nucleus). Rest mass energy of an alpha particle m0c2 =3727.38 MeV.The Heisenberg's uncertainty principle is given as: ∆x∆p=h4π -----(1) where, ∆x is the uncertainity in the position, h is the plank's constant, which is equal to 6.626×10-34 Js. Rearrange and insert the value of ∆x into the formula to determine ∆p. ∆p = h4π ∆x= 6.626×10-34 Js4π×1.56×10-14 m= 3.38×10-21 kg.m/s. Total energy is given as: E=∆pc2+m0c22 -----(2)The first part is calculated as: ∆pc =3.38×10-21 kg.m/s×3×108 m/s=1.014×10-12 J Therefore, ∆pc2=1.082×10-24 J2. The second part is calculated as: m0c2 = 3727.38 MeV= (3727.38×106×1.6×10-19) J= 5.9638×10-10 J Therefore, m0c22= 3.5567×10-19 J2.Insert the above obtained values into the formula for total energy to determine the total energy.Note: 1 eV = 1.6×10-19 J . E = 1.082×10-24 J2+3.5567×10-19 J2=5.9638×10-10 J ×1 eV1.6×10-19 J=3727.3 MeVThe answer is given as follows: The total energy of the particle is equal to 3727.3 Mev.

Answered: 4. The radius of a Uranium-238 nucleus…

4. The radius of a Uranium-238 nucleus is approximately 7.8 × 10-15 m. Using Heisenberg's uncertainty relation, estimate the energy of an a particle that can be emitted from the nucleus, assuming that the a particle was localized inside the nucleus before it emerged. Use sensible units.