At the point of fission, a nucleus of 235U that has 92 protons is divided into two smaller spheres, each of which has 46 protons and a radius of 5.11 ✕ 10-15 m. What is the magnitude of the repulsive force pushing these two spheres apart?
The answer is 3.5*10^3
To calculate the magnitude of the repulsive force pushing the two spheres apart, we can use Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
1. First, we need to calculate the charges of the two smaller spheres. Since each sphere has 46 protons, which each have a charge of +1.6 × 10^-19 C, the charge of each sphere is given by:
q = 46 × (1.6 × 10^-19 C)
2. Next, we calculate the distance between the two spheres. The radius of each sphere, r, is given as 5.11 × 10^-15 m. The total distance between the centers of the two spheres is twice the radius:
d = 2 × r
3. Now that we have the charges and the distance, we can substitute these values into Coulomb's Law:
F = (k * q1 * q2) / (d^2)
In this equation:
F represents the force between the two spheres,
k is the electrostatic constant and equals 8.99 × 10^9 Nm^2/C^2,
q1 and q2 are the charges of the spheres, and
d is the distance between the centers of the spheres.
4. Substituting the given values into the equation:
F = (8.99 × 10^9 Nm^2/C^2) * [ (46 × (1.6 × 10^-19 C)) * (46 × (1.6 × 10^-19 C)) ] / (2 × (5.11 × 10^-15 m))^2
5. Calculating the expression above will provide us with the magnitude of the repulsive force pushing the two spheres apart.
To calculate the magnitude of the repulsive force pushing the two spheres apart, we can use Coulomb's Law, which states that the magnitude of the force between two charged objects is given by:
F = (k * q1 * q2) / r^2
Where:
F is the magnitude of the force
k is the electrostatic constant (9 x 10^9 N m^2/C^2)
q1 and q2 are the charges of the two objects
r is the distance between the centers of the two objects
In this case, the two spheres have the same number of protons, so their charges (q1 and q2) are equal. To find the charge of each sphere, we need to determine the number of protons it contains. Given that a nucleus of 235U has 92 protons, each of the smaller spheres will have 46 protons.
Now, we can determine the distance between the centers of the spheres. The problem states that the radius of each sphere is 5.11 ✕ 10^-15 m. Since the radius represents the distance from the center to the edge of a sphere, the distance between the centers of the spheres (r) will be twice the radius.
r = 2 * 5.11 ✕ 10^-15 m
Now we can substitute these values into Coulomb's Law and calculate the magnitude of the repulsive force:
F = (9 x 10^9 N m^2/C^2) * (46 * 1.6 x 10^-19 C)^2 / (2 * 5.11 ✕ 10^-15 m)^2
Calculating this expression will give us the magnitude of the repulsive force pushing the two spheres apart.
F = k Q1 Q2 / d^2
= 9*10^9 * 1.6*10^-19*1.6*10^-19 / (26.1*10^-30)
= .882 * 10^1 = 8.82 Newtons