To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.
The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,
[tex]A_1*m=M*A_2[/tex]
[tex]A_i =Area[/tex]
M,m = Counts per second
Our radios are given by
[tex]r_1 = 11cm[/tex]
[tex]R_2 = 20cm[/tex]
[tex]m = 65cps[/tex]
Therefore replacing we have that,
[tex]A_1*m=M*A_2[/tex]
[tex]4\pi r_1^2*m = M * 4\pi R_2^2 M[/tex]
[tex]r^2*m=MR^2[/tex]
[tex]M = \frac{m*r^2}{R^2}[/tex]
[tex]M = \frac{65*11^2}{20^2}[/tex]
[tex]M = 19.6625cps[/tex]
Therefore the number of counts expect at a distance of 20 cm is 19.66cps
