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
Answer:
10 m/s
Explanation:
Momentum before collision = momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(8 kg)(8 m/s) + (6 kg)(6 m/s) = (8 kg)(5 m/s) + (6 kg) v
64 kg m/s + 36 kg m/s = 40 kg m/s + (6 kg) v
60 kg m/s = (6 kg) v
v = 10 m/s
Electromagnetic radiation is an energy that is known as light. so electromagnetic radiation will have the same speed as the speed of light which is 3 x 10^8 m/s. so the distance it travel at 55 x 10^-6 s is:
D = ( 3 x 10^8 m/s ) ( 55 x 10^-6 s )
D = 16500 m
Explanation:
The solution is be found in the attachment.