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Found in: Page 1333

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# (a) Calculate the rate at which the Sun generates neutrinos. Assume that energy production is entirely by the proton-proton fusion cycle. (b) At what rate do solar neutrinos reach Earth?

a) The rate at which the Sun generates neutrinos is $1.8×{10}^{38}{s}^{-1}$ .

b) The rate at which the solar neutrinos reach Earth is $8.2×{10}^{28}{s}^{-1}$ .

See the step by step solution

## Step 1: Describe the expression for the rate

The expression for the rate at which the Sun generates neutrinos is given by,

${{\mathbf{R}}}_{{\mathbf{v}}}{\mathbf{=}}\frac{\mathbf{2}\mathbf{P}}{\mathbf{Q}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(}}{\mathbf{1}}{\mathbf{\right)}}$

## Step 2: Find the rate at which the Sun generates neutrinos(a)

The energy per fusion is $Q=26.7MeV$ , and the power of the sun is $P=3.9×{10}^{26}W$ .

Substitute all the known values in equation (1).

${R}_{v}=\frac{2\left(3.9×{10}^{26}W\right)}{26.7MeV}\phantom{\rule{0ex}{0ex}}=\frac{2\left(3.9×{10}^{26}W\right)}{\left(26.7MeV\right)\left(1.602×{10}^{-13}J/MeV\right)}\phantom{\rule{0ex}{0ex}}=1.8×{10}^{38}{s}^{-1}$

Therefore, the rate at which the Sun generates neutrinos is $1.8×{10}^{38}{s}^{-1}$ .

## Step 3: Find the rate at which the solar neutrinos reach Earth(b)

Let ${R}_{v,e}$ be the rate at which the solar neutrinos reach the earth,d is the distance between the earth and the sun and r is the radius of the earth.

The expression of the rate is given by,

${R}_{v,e}={R}_{v}\left(\frac{{\mathrm{\pi r}}^{2}}{4{\mathrm{\pi d}}^{2}}\right)\phantom{\rule{0ex}{0ex}}={R}_{v}\left(\frac{{r}^{2}}{4{d}^{2}}\right).......\left(2\right)$

The radius of the earth is $r=6.4×{10}^{6}m$ , and the distance between the earth and the sun is $d=1.5×{10}^{11}m$ .

Substitute all the known values in equation (2).

${R}_{v,e}=\left(1.8×{10}^{38}{s}^{-1}\right)\left(\frac{{\left(6.4×{10}^{6}m\right)}^{2}}{2{\left(1.5×{10}^{11}m\right)}^{2}}\right)\phantom{\rule{0ex}{0ex}}=8.2×{10}^{28}{s}^{-1}$

Therefore, the rate at which the solar neutrinos reach Earth is $8.2×{10}^{28}{s}^{-1}$ .