A proton is fired from far away toward an iron nucleus. The nucleus has atomic number Z = 26 and a diameter of 9.0 fm. Assuming the nucleus remains at rest, what initial speed is required for the proton to just reach the surface of the nucleus? (Use conservation of energy; k = 8.99×10^9 N·m^2/C^2; e = 1.602×10^-19 C; m_p = 1.673×10^-27 kg.)

• A. 4.0×10^7 m/s
• B. 1.0×10^8 m/s
• C. 8.9×10^7 m/s
• D. 2.0×10^8 m/s

Answer: (C)

Conservation of energy between the proton’s initial kinetic energy and the electrostatic repulsion from the iron nucleus sets the speed needed to just touch the surface. As the proton approaches, the Coulomb potential grows; to reach the nuclear surface its initial kinetic energy must equal the Coulomb energy at that surface.

The distance to use is the nuclear radius, since “surface” means the proton’s center is at about R = 4.5 fm = 4.5×10^-15 m from the nucleus center. The Coulomb potential energy is U = k Z e^2 / R, with k = 8.99×10^9 N·m^2/C^2, Z = 26, e = 1.602×10^-19 C.

Compute U: U ≈ [8.99×10^9 × 26 × (1.602×10^-19)^2] / (4.5×10^-15) ≈ 1.33×10^-12 J.

Set the initial kinetic energy equal to this: (1/2) m_p v^2 = 1.33×10^-12 J, with m_p = 1.673×10^-27 kg. Solve for v:

v^2 = (2×1.33×10^-12) / (1.673×10^-27) ≈ 1.59×10^15, so v ≈ 3.99×10^7 m/s.

The proton must arrive at about 4×10^7 m/s to just reach the surface.