An alpha particle (which has two protons) is sent directly toward a target nucleus containing 80 protons. The alpha particle has an initial kinetic energy of 0.79 pJ. What is the least center-to-center distance the alpha particle will be from the target nucleus, assuming the nucleus does not move? The charge of a proton is 1.602 x 10-19 C.

He has the right equation, but the wrong exponents.

Equation = (r = (KQ)/K0)

K = 9 * 10^9
Q = (2*e*#protons*e)
K0 = initial ke * 10^-12

Example:
(K*(2*1.602E-19*86*1.602E-19))/.39E-12
= 1.017E-13m

The potential energy due to the Coulomb replsion force is

-k Q1 Q2/R
at separation distance.
k is the Coulomb constant and Q1 and Q2 are the two positive nuclear charges. When the nuclei are at closest separation distance d, the relative velocity is zero. You can ignore the effect of the 80 electrons around the larger nucleus.

Initial kinetic energy = potential energy change
0.79*10^-12 J = k Q1 Q2/d = 2*80 k e^2/d

Solve for d

To find the least center-to-center distance between the alpha particle and the target nucleus, we need to use the concept of electrostatic potential energy.

The electrostatic potential energy between two charged particles is given by the equation:

U = (k * q1 * q2) / r

Where U is the potential energy, k is the electrostatic constant (9 x 10^9 N m^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between their centers.

In this case, the alpha particle has two protons with a total charge of +2e (where e is the elementary charge, 1.602 x 10^-19 C). The target nucleus has 80 protons with a charge of +80e.

The initial kinetic energy of the alpha particle is given as 0.79 pJ (pJ means picojoules, 10^-12 J).

To find the least center-to-center distance, we can equate the initial kinetic energy to the potential energy at this distance:

0.79 pJ = (k * (2e) * (+80e)) / r

Rearranging the equation to solve for r:

r = (k * (2e) * (+80e)) / (0.79 pJ)

Plugging in the values:

r = (9 x 10^9 N m^2/C^2) * (2 * (1.602 x 10^-19 C) * (80 * (1.602 x 10^-19 C))) / (0.79 x 10^-12 J)

Calculating the value gives us:

r = 0.36 x 10^-12 m

Therefore, the least center-to-center distance between the alpha particle and the target nucleus is approximately 0.36 picometers.

To find the least center-to-center distance between the alpha particle and the target nucleus, we can use the concept of energy conservation. We'll assume that the alpha particle starts from rest, and the only force acting on it is the electrostatic force between the alpha particle and the nucleus.

First, let's convert the initial kinetic energy of the alpha particle from picojoules (pJ) to joules (J):

0.79 pJ = 0.79 x 10^-12 J

Next, we can calculate the electrostatic potential energy (U) between the alpha particle and the nucleus using Coulomb's law:

U = (k * q1 * q2) / r

Where:
- k is the electrostatic constant (8.99 x 10^9 Nm^2/C^2)
- q1 and q2 are the charges of the alpha particle and the target nucleus, respectively
- r is the distance between them

Since the alpha particle has two protons, its charge can be calculated as:

q1 = 2 * (1.602 x 10^-19 C) = 3.204 x 10^-19 C

The charge of the target nucleus is given as 80 protons, so:

q2 = 80 * (1.602 x 10^-19 C) = 1.2816 x 10^-17 C

Now, we can equate the initial kinetic energy to the electrostatic potential energy:

0.79 x 10^-12 J = (k * q1 * q2) / r

To find the least center-to-center distance, we need to rearrange the equation and solve for r:

r = (k * q1 * q2) / (0.79 x 10^-12 J)

Substituting the values:

r = (8.99 x 10^9 Nm^2/C^2 * 3.204 x 10^-19 C * 1.2816 x 10^-17 C) / (0.79 x 10^-12 J)

Calculating this expression will give us the least center-to-center distance between the alpha particle and the target nucleus.