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A proton is fired at 2.10×107 m/s toward a mercury atom from far away. Mercury has a diameter of 14.0 fm and is element number 80. What is the closest distance the proton will get to the surface of the mercury atom?

Question

A proton is fired at 2.10×107 m/s toward a mercury atom from far away. Mercury has a diameter of 14.0 fm and is element number 80. What is the closest distance the proton will get to the surface of the mercury atom?

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Solution

To solve this problem, we need to use the formula for the distance of closest approach, which is derived from the conservation of energy and the concept of electrostatic force. The formula is:

d = k * Ze^2 / (2KE)

where:

  • d is the distance of closest approach
  • k is Coulomb's constant (8.99 x 10^9 N m^2/C^2)
  • Z is the atomic number of the target atom (for mercury, Z = 80)
  • e is the elementary charge (1.60 x 10^-19 C)
  • KE is the kinetic energy of the proton

First, we need to calculate the kinetic energy (KE) of the proton. The formula for kinetic energy is:

KE = 1/2 * m * v^2

where:

  • m is the mass of the proton (1.67 x 10^-27 kg)
  • v is the velocity of the proton (2.10 x 10^7 m/s)

Substituting the values, we get:

KE = 1/2 * (1.67 x 10^-27 kg) * (2.10 x 10^7 m/s)^2 KE = 3.50 x 10^-14 J

Now, we can substitute the values into the formula for the distance of closest approach:

d = (8.99 x 10^9 N m^2/C^2) * 80 * (1.60 x 10^-19 C)^2 / (2 * 3.50 x 10^-14 J) d = 2.05 x 10^-14 m

However, the question asks for the closest distance the proton will get to the surface of the mercury atom, not the center. So, we need to subtract the radius of the mercury atom from this distance. The diameter of the mercury atom is given as 14.0 fm (femtometers), so the radius is 7.0 fm, or 7.0 x 10^-15 m.

Therefore, the closest distance the proton will get to the surface of the mercury atom is:

d - radius = 2.05 x 10^-14 m - 7.0 x 10^-15 m = 1.35 x 10^-14 m.

This problem has been solved

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