Answer:
16 cm
Explanation:
For protons:
Energy, E = 300 keV
radius of orbit, r1 = 16 cm
the relation for the energy and velocity is given by

So,
.... (1)
Now,

Substitute the value of v from equation (1), we get

Let the radius of the alpha particle is r2.
For proton
So,
... (2)
Where, m1 is the mass of proton, q1 is the charge of proton
For alpha particle
So,
... (3)
Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle
Divide equation (2) by equation (3), we get

q1 = q
q2 = 2q
m1 = m
m2 = 4m
By substituting the values

So, r2 = r1 = 16 cm
Thus, the radius of the alpha particle is 16 cm.
Answer:
Hey
It would have to be C because no net energy is lost.
The answer to your question is A
A proton is held at rest in a uniform electric field. When it is released, the proton will lose its kinetic energy.
Kinetic energy
The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes. The body exerts the same amount of effort when slowing down from its current pace to a condition of rest. Formally, kinetic energy is any term that includes a derivative with respect to time in the Lagrangian of a system.
To learn more about kinetic energy refer here:
brainly.com/question/11301578
#SPJ4
The correct answer would be to express large and small numbers.