Respuesta :
Answer:
we know our detector is limited to a maximum count rate of 2000 particles/s, and we would expect the count of 4252 particles/s as angle of 6 degree. That means we would have to modify our detector if we want to measure proper count number.
Explanation:
To get the number of particle scattered per unit area, N(Ф) into the ring at the scattering angle Ф is
Ф = tet
N(tet) = [tex]\frac{Ni*nt}{16} * (\frac{e^{2} }{4*pi*e0})^{2} * \frac{Z^{2}_{1} * Z^{2}_{2} }{r^{2}*K^{2} *sin^{4} (tet/2) }[/tex]
We can first calculate ratio of scattered particles at 50 and 6 degree angles using the equation
[tex]\frac{N(50)}{N(6)} = \frac{sin^{4} (3) }{sin^{4} (25) } \\\frac{N(50)}{N(6)} = 2.35 * 10^{-4}[/tex]
from there we can calculate number of scattered particles at 6 degree
[tex]N(6) = \frac{N(50)}{2.35*10^{-4} } \\N(6) = \frac{1 particles/seconds}{2.35*10^{-4} } \\N(6) = 4252 particles/seconds[/tex]
we know our detector is limited to a maximum count rate of 2000 particles/s, and we would expect the count of 4252 particles/s as angle of 6 degree. That means we would have to modify our detector if we want to measure proper count number.
