Question

Alpha particles of kinetic energy 6.4 MeV are incident at a rate of 3.2 107 per second on a gold foil of thickness 3.2 10-6 m. A circular detector of diameter 1.0 cm is placed 13 cm from the foil at an angle of 30° with the direction of the incident alpha particles. At what rate does the detector measure scattered alpha particles? (The molar mass of gold is 197.0 g/mol and its density is 19.3 g/cm3.)

Alpha particles of kinetic energy 6.4 MeV are incident at a rate of 3.2 x 107 per second on a gold foil of thickness 3.2 x 10-6 m. A. circular detector of diameter 1.0 cm is placed 13 cm from the foil at an angle of 30° with the direction of the incident alpha particles. At what rate does the detector measure scattered alpha particles? (The molar mass of gold is 197.0 g/mol and its density is 19.3 g/cm3.) s-1 Additional Materials eBook

Answer 1

Expression for Rutherford scattering cross-section is, Relation between number of moles (n), avogadro number (NA). molecular weight(M)and density() is »END

„ 6.023 - 102 19.3 7 = 197

n = 5.90 * 1023 atom/m

Here, 410 Here, J=1.44 MeV.fm Substitute all the data in the main equation solve for N(O).

ਵT0) · " (rt 6 ) , 01 0 069

= 279.28 + 102 #315.95 + 10-30 + 222.85

= 0.1966 m2

Probability of alpha particles strikes foil is, Pr) = N(O) A

= 0.10663- (ཀ (1-10-)

= 1.54 + 10-6

Rate of alpha particles strkes the detector is,

r(t) = 1.544 * 10-6 * 3.2 * 107

r(t) = 49.40 s-1