An alpha particle(charge +2e) is sent at a speed towards a gold nucleus( +79e ).
1) what is the electric force acting on the alpha particle when it is 2.0 x 10^-14 m from the gold nucleus.
2) if the alpha particle came to a stop at this position after traveling 8.22 x 10^-13, what was the initial velocity of the particle. (mass= 6.64 x 10^-27 kg)
f = k Q1 Q2 / d^2
change in Ke = 1/2 m v^2
= change in U = k Q1 Q2 (1/r2 -1/r1)
r1 = 2*10^-14 + 8.22*10^-13
r2 = 2*10^-14
To solve both questions, we need to use Coulomb's law to determine the electric force between the alpha particle and the gold nucleus. Coulomb's law states that the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
1) To find the electric force acting on the alpha particle, we can use the formula:
F = k * (q1 * q2) / r^2
Where:
F is the electric force
k is Coulomb's constant, approximately 9 x 10^9 Nm^2/C^2
q1 and q2 are the charges of the two objects
r is the distance between the two objects
In our case:
q1 = +2e (2 times the elementary charge, 1.6 x 10^-19 C)
q2 = +79e (79 times the elementary charge)
r = 2.0 x 10^-14 m
Plugging in the values, we get:
F = (9 x 10^9 Nm^2/C^2) * [(2 * 1.6 x 10^-19 C) * (79 * 1.6 x 10^-19 C)] / (2.0 x 10^-14 m)^2
Solving this equation will give you the electric force acting on the alpha particle when it is 2.0 x 10^-14 m from the gold nucleus.
2) To find the initial velocity of the alpha particle, we can use the principle of conservation of energy. The alpha particle initially has kinetic energy, and as it approaches the gold nucleus, it loses kinetic energy and comes to a stop.
We can equate the initial kinetic energy of the alpha particle to the amount of work done by the electric force in bringing it to a stop. The work done by the electric force can be calculated using the formula:
Work = electric force * distance
Since the alpha particle comes to a stop, its final kinetic energy is zero. Therefore, the initial kinetic energy is equal to the work done:
(1/2) * m * (initial velocity)^2 = electric force * distance
Plugging in the mass (m = 6.64 x 10^-27 kg), electric force (from part 1), and distance (8.22 x 10^-13 m), we can solve this equation to find the initial velocity of the alpha particle.
Remember to use the appropriate units throughout the calculations.
To find the electric force acting on the alpha particle when it is 2.0 x 10^-14 m from the gold nucleus, you can use Coulomb's law:
1) Coulomb's law states that the electric force between two charged objects is given by the equation:
F = (k * |q1 * q2|) / r^2
where F is the force, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q1 and q2 are the charges of the objects (in this case, the alpha particle has a charge of +2e and the gold nucleus has a charge of +79e), and r is the distance between the centers of the two charged objects.
Plugging in the values, we have:
F = (8.99 x 10^9 N m^2/C^2) * (2e * 79e) / (2.0 x 10^-14 m)^2
Now, simplifying the expression:
F = (8.99 x 10^9 N m^2/C^2) * (2 * 79 * (1.6 x 10^-19 C)^2) / (2.0 x 10^-14 m)^2
F = 180.835 N
To find the initial velocity of the particle when it comes to a stop at a position 8.22 x 10^-13 m away from the gold nucleus, you can use the work-energy theorem:
2) The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the initial kinetic energy of the alpha particle is equal to the work done by the electric force to bring it to a stop.
The work done by the electric force is given by:
Work = F * d
where F is the force and d is the displacement. In this case, the force is the electric force calculated earlier, and the displacement is the distance the alpha particle traveled before coming to a stop.
Work = 180.835 N * 8.22 x 10^-13 m
Now, the work done is equal to the change in kinetic energy:
Work = ΔKE
ΔKE = (1/2) * m * (v^2 - u^2)
where m is the mass of the alpha particle, v is its final velocity (which is 0 m/s at the point it comes to a stop), and u is its initial velocity.
Since we know the values of ΔKE, m, and u, we can solve for v:
(1/2) * m * (v^2 - u^2) = 180.835 N * 8.22 x 10^-13 m
(1/2) * (6.64 x 10^-27 kg) * (0 - u^2) = 180.835 N * 8.22 x 10^-13 m
-0.5 * (6.64 x 10^-27 kg) * u^2 = 180.835 N * 8.22 x 10^-13 m
u^2 = (2 * 180.835 N * 8.22 x 10^-13 m) / (6.64 x 10^-27 kg)
Now, we can solve for u:
u = sqrt((2 * 180.835 N * 8.22 x 10^-13 m) / (6.64 x 10^-27 kg))