Question

An alpha particle (4He) undergoes an elastic collision with a stationary uranium nucleus (235U). What percent of the kinetic energy of the alpha particle is transferred to the uranium nucleus? Assume the collision is one-dimensional.

a) 0%
b) 25%
c) 50%
d) 100%

Answer 1

The percent of kinetic energy transferred to the uranium nucleus from the alpha particle in a one-dimensional elastic collision, based on their mass ratio, is approximately 3.3%, which is not explicitly given among the options.

Step-by-step explanation:

The question deals with a one-dimensional elastic collision between an alpha particle and a uranium nucleus. In elastic collisions, the total kinetic energy is conserved, but how this energy is distributed between the two colliding objects depends on their masses.

Using the conservation of momentum and kinetic energy principles, the fraction of kinetic energy transferred from the alpha particle to the uranium nucleus can be determined. Since the mass of the alpha particle (4He) is much smaller than that of the uranium nucleus (235U), and given their mass ratio, we can apply the conservation of kinetic energy equation for elastic collisions to find the percentage of kinetic energy transferred.

The formula for the kinetic energy transferred in an elastic collision between two objects where one is initially stationary is K.E. transferred = (4*m1*m2)/(m1+m2)² where m1 and m2 are the masses of the two objects. In this case, m1 is the alpha particle and m2 is the uranium nucleus. Plugging in the mass ratio of 4 to 235 yields approximately 0.033 or 3.3%, which is not one of the options provided in the question.