Question

The differential cross section for scattering 6.5-MeV alpha particles at 120° off a silver nucleus is about 0.5 barns/sr. If a total of 10 10 alphas impinge on a silver foil of thickness 1 itm and if we detect the scattered particles using a counter of area 0.1 mm 2 at 120° and 1 cm from the target, about how many scattered alphas should we expect to count? (Silver has a specific gravity of 10.5, and atomic mass of 108.) 14.12 ***

Answer 1

Nsc=30

Step-by-step explanation:

The solid angle subtended by the counter is

d=0.1mm2/(1cm)2=(10)3 sr

d=(10)10x10.5gcm3x10-4cm/108x1.6610x10-24gx0.510-24cm2x10-3=30