The oxygen nucleus 16O has a radius of 3.0 fm. With what speed must a proton be fired towards an oxygen nucleus to have a turning point of 1.0 fm from the surface?

To determine the speed at which a proton must be fired towards an oxygen nucleus (16O) to reach a turning point of 1.0 femtometer (fm) from its surface, we can use concepts from nuclear physics and classical mechanics. The turning point indicates the closest distance the proton can approach the nucleus before being repelled due to the electromagnetic force. Let's break this down step by step.

Understanding the Problem

The radius of the oxygen nucleus is given as 3.0 fm. Therefore, the distance from the center of the nucleus to the surface is 3.0 fm. If the proton is to have a turning point of 1.0 fm from the surface, we need to calculate the total distance from the center of the nucleus to this turning point.

Calculating the Total Distance

The total distance from the center of the nucleus to the turning point can be calculated as follows:

• Radius of the nucleus: 3.0 fm
• Turning point distance from the surface: 1.0 fm

Thus, the total distance (d) from the center of the nucleus to the turning point is:

d = Radius of nucleus + Turning point distance = 3.0 fm + 1.0 fm = 4.0 fm

Applying Energy Conservation

To find the speed of the proton, we can apply the principle of conservation of energy. The kinetic energy of the proton when it is fired towards the nucleus will be converted into potential energy at the turning point. The potential energy (U) between the proton and the nucleus can be expressed using the formula for the electrostatic potential energy:

U = k * (Z1 * Z2 * e^2) / r

Where:

• k is Coulomb's constant (approximately 8.99 x 10^9 N m²/C²)
• Z1 and Z2 are the atomic numbers of the interacting particles (for a proton and oxygen, Z1 = 1 and Z2 = 8)
• e is the elementary charge (approximately 1.6 x 10^-19 C)
• r is the distance between the charges (4.0 fm = 4.0 x 10^-15 m)

Calculating Potential Energy

Substituting the values into the potential energy formula:

U = (8.99 x 10^9 N m²/C²) * (1 * 8 * (1.6 x 10^-19 C)²) / (4.0 x 10^-15 m)

Calculating this gives:

U ≈ (8.99 x 10^9) * (8 * 2.56 x 10^-38) / (4.0 x 10^-15)

U ≈ (8.99 x 10^9) * (20.48 x 10^-38) / (4.0 x 10^-15)

U ≈ 4.59 x 10^-13 J

Relating Kinetic Energy to Potential Energy

The kinetic energy (K) of the proton when it is fired can be expressed as:

K = (1/2) * m * v²

Where:

• m is the mass of the proton (approximately 1.67 x 10^-27 kg)
• v is the speed of the proton

At the turning point, the kinetic energy will equal the potential energy:

(1/2) * m * v² = U

Substituting the values we have:

(1/2) * (1.67 x 10^-27 kg) * v² = 4.59 x 10^-13 J

Solving for Speed

Now, we can solve for v:

v² = (2 * 4.59 x 10^-13 J) / (1.67 x 10^-27 kg)

v² ≈ (9.18 x 10^-13) / (1.67 x 10^-27)

v² ≈ 5.49 x 10^14

Taking the square root gives:

v ≈ 2.34 x 10^7 m/s

Final Result

Therefore, the speed at which the proton must be fired towards the oxygen nucleus to reach a turning point of 1.0 fm from its surface is approximately 2.34 x 10^7 m/s.