Answer:
Explanation:
The volume of the tank = 50 kton
50 kton = 5 × 10⁷ kg
Since 18 grams of water will contain: 10 electrons × 6.023 × 10²³
Then;
5× 10⁷ kg will contain [tex]( \dfrac{5 \times 10^7 \times 10^3}{18}) \times 10 \times 6.023 \times 10^{23}[/tex]
= 1.67 × 10³⁴ electrons
(b)
Suppose:
[tex]f_{neutrino} = \dfrac{2f_o}{26.2 MeV} = 6.7\times 10^{10} \ s^{-1} cm^{-2}[/tex]
Then;
10⁻⁶ of [tex]f_{neutrino} = 6.7 \times 10^{10} \times 10^{-6} \ s^{-1} cm^{-2}[/tex]
[tex]=6.7 \times 10^{4}\ s^{-1} cm^{-2}[/tex]
Thus, the number of high energy neutrinos which will interact with water is:
= [tex]6.7 \times 10^4 \times \sigma[/tex]
= [tex]6.7 \times 10^4 \times 10^{-43}[/tex]
= [tex]6.7 \times 10^{-39} s^{-1}[/tex]
For 1.67 × 10³⁴ electrons, the detection rate is:
[tex]6.7 \times 10^{-39} \times 1.67 \times 10^{34}[/tex]
[tex]= 11.19 \times 10^{-5} \ s^{-1}[/tex]
= 9.668 per day
