calculate the distance of closest approach for a head-on collision between a 5.30 mev alpha particle and the nucleus of a copper atom.

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19-12-2022
Chemistry

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calculate the distance of closest approach for a head-on collision between a 5.30 mev alpha particle and the nucleus of a copper atom.

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Papreso Papreso · Answer 1 · 2022-12-19T13:02:12-05:00

To calculate the distance of closest approach for a head-on collision between an alpha particle and the nucleus of a copper atom, we can use the formula:

r = Z1Z2e^2 / (2m1v^2)

where r is the distance of closest approach, Z1 and Z2 are the atomic numbers of the alpha particle and the copper nucleus, respectively, e is the charge of the electron (1.6 x 10^-19 C), m1 is the mass of the alpha particle, and v is the velocity of the alpha particle.

The atomic number of an alpha particle is 2, since it consists of two protons and two neutrons. The atomic number of copper is 29. The mass of an alpha particle is 6.644 x 10^-27 kg, and its velocity can be calculated from its kinetic energy using the formula:

v = sqrt(2*KE/m1)

where KE is the kinetic energy of the alpha particle in joules. The kinetic energy of a 5.30 MeV alpha particle is 5.30 x 10^6 eV, or 8.478 x 10^-13 J.

Substituting these values into the formula for the distance of closest approach, we get:

r = 229(1.6 x 10^-19 C)^2 / (2*(6.644 x 10^-27 kg)(sqrt(28.478 x 10^-13 J/6.644 x 10^-27 kg))^2

= 5.936 x 10^-15 m

Therefore, the distance of closest approach for a head-on collision between a 5.30 MeV alpha particle and the nucleus of a copper atom is approximately 5.936 x 10^-15 meters.