A beam of 8.3 Mev alpha particles is directed at an aluminum foil. it is found that the Rutherford scattering formula cease to be obeyed at scattering angles exceedin about 60 .if alpha particle is assumed to have a radius 2*10^-15m. find the radius of the aluminum nucleus

To find the radius of the aluminum nucleus, we can use the Rutherford scattering formula.

The Rutherford scattering formula relates the scattering angle (θ) to the impact parameter (b), the distance of closest approach between the alpha particle and the nucleus. It is given by:

θ = (2Ze^2 / 4πε₀E) * (1 / b)

where:
θ is the scattering angle,
Z is the atomic number of the target nucleus,
e is the elementary charge (1.602 x 10^-19 C),
ε₀ is the permittivity of free space (8.854 x 10^-12 C²/N m²),
E is the kinetic energy of the alpha particle, and
b is the impact parameter.

In this case, we have 8.3 MeV (or 8.3 x 10^6 eV) alpha particles directed at an aluminum foil. Let's assume aluminum has an atomic number of Z=13.

First, convert the alpha particle energy to joules:
E = 8.3 x 10^6 eV * (1.602 x 10^-19 J/eV) = 1.329 x 10^-12 J

Next, we need to find the scattering angle where the Rutherford scattering formula ceases to be obeyed, which is 60 degrees (or 60 * π/180 radians).

Since the formula becomes invalid beyond this angle, we can rearrange the formula and solve for the impact parameter (b):

b = (2Ze^2 / 4πε₀E) * (1 / θ)

b = (2 * 13 * (1.602 x 10^-19 C)^2 / (4 * π * 8.854 x 10^-12 C²/N m² * 1.329 x 10^-12 J)) * (1 / (60 * π/180))

b = 4.47 x 10^-14 m

Now, we can calculate the radius of the aluminum nucleus by subtracting the radius of the alpha particle (given as 2 x 10^-15 m) from the impact parameter:

Radius of the aluminum nucleus = b - Radius of alpha particle
= 4.47 x 10^-14 m - 2 x 10^-15 m
= 4.27 x 10^-14 m

Therefore, the radius of the aluminum nucleus is approximately 4.27 x 10^-14 meters.