A beam of 3 MeV α particles scatters from a thin gold foil. What is the ratio of the number of α particles scattered to angles greater than 4° to the number scattered to angles greater than 16°?

Ratio of number of alpha particles is 16.20

Explanation:

In Rutherford Scattering, the fraction of incident beam particles scattered with angle θ and grater than θ is given by the relation :

Here Z₁ is the atomic number of target, Z₂ is atomic number of incident particles, e is electronic charge, K is kinetic energy of incident particles, n is target particle density and t is thickness of the target foil.

According to the problem, number of alpha scattered to angles greater than 4° is given by the relation :

     ....(1)

Number of alpha scattered to angles greater than 16° is given by the relation :

   .....(2)

Ratio of equation (1) and (2) gives :

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s