Answer:
The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is
8.00 x 10-13J
Explanation:
In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).
Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)
Although, the kinetic energy is converted to potential energy in Coulomb's law equation.
That is,
1/2(mv^2) = (K* q1q2)/r
Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter
Average speed = (total distance covered) / (time to cover the distance)
Total distance = (77km + 66km) = 143 kilometers
Time to cover the distance = 2 hours
Average speed = (143 km) / (2 hours) = 71.5 km per hour
D is the wrong answer. New information does often completely change the theory. Its hard to change something and leave the major theory intact.
Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s
no, work is = force * distance or displacement
