An alpha particle has a kinetic energy of about 5 MeV after they are far from the nucleus. Use this information to estimate the size of a nucleu by assuming that the alpha particle leaves the edge of the nucleus with zero kinetic energy and that the electrostatic potential energy is the only potential energy it has at that position. Assume the charge on the nucleus is +88e. The charge on an alpha particle is +2e while the mass of an alpha particle is 4.0026 amu where 1 amu=1.66x10^27 kg.
k*Q1*Q2/R = 5 MeV = potential energy change from R to infinity.
Convert the 5 MeV to Joules and solve for R
k is the Coulomb constant
Q1 = 2e
Q2 = 88e
e = electron charge
To estimate the size of a nucleus using the given information, we can apply the conservation of mechanical energy principle. The initial kinetic energy of the alpha particle is converted into its final electrostatic potential energy at the edge of the nucleus.
We are given that the alpha particle has a kinetic energy of about 5 MeV (5x10^6 eV) after being far from the nucleus. The charge on the nucleus is +88e where e is the elementary charge (1.6x10^-19 C). The charge on an alpha particle is +2e. The mass of an alpha particle is 4.0026 amu, which is equivalent to 4.0026 x 1.66x10^-27 kg (since 1 amu = 1.66x10^-27 kg).
We can now proceed to calculate the electrostatic potential energy of the alpha particle at the edge of the nucleus using the formula:
Potential energy = (Electrostatic potential energy constant) x ((Charge of the alpha particle) x (Charge of the nucleus)) / (Radius of the nucleus)
Substituting the given values:
Potential energy = (8.99x10^9 Nm^2/C^2) x ((2e) x (+88e)) / (Radius of the nucleus)
Since the alpha particle leaves the edge of the nucleus with zero kinetic energy, its total mechanical energy is zero at the edge of the nucleus. Thus, we have the equation:
Total mechanical energy = Kinetic energy + Potential energy
0 = 5 MeV + (8.99x10^9 Nm^2/C^2) x ((2e) x (+88e)) / (Radius of the nucleus)
Now, we can solve this equation for the radius of the nucleus.
Rearranging the equation:
Radius of the nucleus = ((8.99x10^9 Nm^2/C^2) x ((2e) x (+88e)) / (-5 MeV)
Radius of the nucleus ≈ ((8.99x10^9 Nm^2/C^2) x ((2e) x (+88e)) / (-5x10^6 eV)
Using the appropriate conversion factors:
Radius of the nucleus ≈ ((8.99x10^9 Nm^2/C^2) x ((2 x 1.6x10^-19 C) x (88 x 1.6x10^-19 C)) / (-5x10^6 x 1.6x10^-19 J)
After performing the calculations, we can determine the approximate value for the size of the nucleus using the given information and the equations mentioned above.