Question

Determine the range (in mg/cm^2 and in cm) of a 12.0 MeV proton in air and in Cu.

Determine the range (in mg/cm^2 and in cm) of a 12.0 MeV alpha particle in air and water.

Estimate the tinckness of Al (in mg/cm^2 and in cm) needed to reduce the energy of a 320 MeV O-16 ion to an energy of 100 MeV

Answer 1

Range(R) in air in cm is defined by the mathematical formula is = 0.31*E^(3/2), where E is the energy of alpha particle. So, R(cm)= 0.31*(12)^(3/2) = 12.886 as E= 12 Mev. Similarly, range in air in mg/cm^2 is given by R(mg/cm^2) = 0.40 * E^(3/2) = 16.627. Range of proton in mg/ cm^2 of energy 12 Mev is defined by Rz= 0.173*E^(3/2)*A^(1/3) =0.173*(12)^(3/2)*1^(1/3) as Atomic mass = A = 1 for H so range = 7.191 mg/ cm^2 , In Cu in mg/ cm^2 the range is = 0.173* (12)^(3/2)* (63.5)^(1/3) = 28.690. The value of range in cm can be obtained by dividing the range in mg/ cm^2 by density of absorber in mg/ cm^3 . Here the density if given then we can calculate.