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:
i don't understand the hw
Kinetic energy is the energy possessed by an object on motion. it is expressed as follows:
KE = 0.5mv^2
where m is the mass and v is the velocity of the object. We calculate as follows:
KE = 0.5mv^2
1.1x10^9 J = 0.5(8.0x10^4 kg) v^2
v = 165.83 m/s
47m total just add them up 35 + 12 = 47
The answer is D since work done must have displacement..if theres force but no displacement there will be no work done based on W=Fs
