What kinetic energy, in MeV, must a proton be fired toward a lead nucleus to have a turning point 8.2 fm from the surface? The lead nucleus 207 Pb 82 has a radius of 7.1 fm. Ignore relativistic effects. Please show work.

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The  lead atom `^207_82 Pb` has a mass of 207 and the atomic number is 82. There are 82 protons in the atom's nucleus. The radius of the nucleus is 7.1 fm.

There is a force of repulsion between two bodies that are positively charged. When a proton is fired at the nucleus, work is done that allows the proton to approach the nucleus. The potential energy of the proton at distance d from the nucleus is U = k*Q*q/r, where r is the distance of the proton from the center of the nucleus, Q is the charge in the nucleus and q is the charge of the proton. If the proton has an initial kinetic energy K, at the turning point, this is equal to the potential energy given by k*Q*q/r.

k is Coulomb's constant equal to 9*10^9 N*m^2/C^2. The charge of a proton is 1.602*10^-19 C. At a distance of  8.2 fm from the surface of the nucleus, the potential energy of the proton is:


= 1.23*10^-12 J

One joule is equal to 6.242*10^18 eV.

1.23*10^-12 J is equal to 7.72*10^18 eV.

The kinetic energy of the proton should be 7.72*10^18 eV for it to have the required turning radius.

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