A 5.0 MeV α-particle approaches a gold nucleus head-on. Take the atomic number of gold as 79 and use the electrostatic constant in nuclear units, e² ÷ (4πε₀) = 1.44 MeV·fm. Neglect nuclear recoil. If the distance of closest approach is approximately N fm, find N to the nearest integer.
Physics · Atomic Structure and X-rays · atomic structure and bohr atomic model
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Question

A 5.0 MeV α-particle approaches a gold nucleus head-on. Take the atomic number of gold as 79 and use the electrostatic constant in nuclear units, e² ÷ (4πε₀) = 1.44 MeV·fm. Neglect nuclear recoil. If the distance of closest approach is approximately N fm, find N to the nearest integer.

Physics · Atomic Structure and X-rays · atomic structure and bohr atomic model

G1 Physics · Accepted Answer

At the distance of closest approach, the α-particle momentarily comes to rest. Its initial kinetic energy is converted into electrostatic potential energy. Charge on α-particle = +2e. Charge on gold nucleus = +79e. Therefore, K = [2×79×1.44] ÷ r(min), where K is in MeV and r(min) is in fm. Substitute K = 5.0 MeV: r(min) = 2×79×1.44 ÷ 5.0 = 227.52 ÷ 5.0 = 45.504 fm. Nearest integer: N = 46.