Question

On Earth, the number flux of solar neutrinos from the p-p chain is:

Answer 1

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 $( \dfrac{5 \times 10^7 \times 10^3}{18}) \times 10 \times 6.023 \times 10^{23}$

= 1.67 × 10³⁴ electrons

(b)

Suppose:

$f_{neutrino} = \dfrac{2f_o}{26.2 MeV} = 6.7\times 10^{10} \ s^{-1} cm^{-2}$

Then;

10⁻⁶ of $f_{neutrino} = 6.7 \times 10^{10} \times 10^{-6} \ s^{-1} cm^{-2}$

$=6.7 \times 10^{4}\ s^{-1} cm^{-2}$

Thus, the number of high energy neutrinos which will interact with water is:

= $6.7 \times 10^4 \times \sigma$

= $6.7 \times 10^4 \times 10^{-43}$

= $6.7 \times 10^{-39} s^{-1}$

For  1.67 × 10³⁴ electrons, the detection rate is:

$6.7 \times 10^{-39} \times 1.67 \times 10^{34}$

$= 11.19 \times 10^{-5} \ s^{-1}$

= 9.668 per day