Question

If a neutrino interacted with a quark every time their separation was within the ${\mathbf{10}}^{\mathbf{-}\mathbf{18}}\mathbf{\text{\hspace{0.17em}m}}$ range generally accepted for the weak force, then the cross-section of a neutron or proton “seen” by a neutrino would be on the order of ${\mathbf{10}}^{\mathbf{-}\mathbf{36}}{\mathbf{\text{\hspace{0.17em}m}}}^{\mathbf{2}}$. Even at such separation, however the probability of interactions is quite small. The nucleon appears to have an effective cross-section of only about ${\mathbf{10}}^{\mathbf{-}\mathbf{48}}{\mathbf{\text{\hspace{0.17em}m}}}^{\mathbf{2}}$.

(a) About how many nucleons are there in a column through the earth’s center of 1 m² cross-sectional area?

(b) what is the probability that a given neutrino passing through space and encountering earth will actually “hit”?

Short answer

(a) The resultant answer is$4\times {10}^{37}$nucleons/m^2.

(b) The probability is $4\times {10}^{-11}$.

Step-by-step solution

Given data

Cross-sectional area $=1{\text{\hspace{0.17em}m}}^2$.

Concept of Density

Density: mass of a unit volume of a material substance.

The formula for density is $\mathbf{d}\mathbf{=}\mathbf{M}\mathbf{/}\mathbf{V}$ ,

where $\mathbf{d}$ is density, $\mathbf{M}$ is mass, and $\mathbf{V}$ is volume.

Density is commonly expressed in units of grams per cubic centimeter.

Calculate the volume 

(a)

Substitute $5.98\times {10}^{24}\text{kg}$for $M_E$and $6.37\times {10}^6\text{m}$ for $R_E$ as:

$\begin{array}{l}{\mathrm{\rho }}_{\mathrm{E}}=\frac{(5.98\times {10}^{24}\text{\hspace{0.17em}kg})}{\frac{4}{3}\mathrm{\pi }{(6.37\times {10}^6\text{\hspace{0.17em}m})}^3}\\ {\mathrm{\rho }}_{\mathrm{E}}=5.5\times {10}^3{\text{\hspace{0.17em}kg/m}}^{\text{3}}\end{array}$

Thus, the density of Earth in $5.5\times {10}^3\text{kg}/{\text{m}}^3$.

Now calculate the volume of this 1 square meter of the earth all across the center as:

$\begin{array}{l}\left(2{\mathrm{R}}_{\mathrm{E}}\right)(1{\text{\hspace{0.17em}m}}^3)=2(6.37\times {10}^6\text{\hspace{0.17em}m})\left(1{\text{\hspace{0.17em}m}}^2\right)\\ \left(2{\mathrm{R}}_{\mathrm{E}}\right)(1{\text{m}}^3)=1.27\times {10}^7{\text{\hspace{0.17em}m}}^3\end{array}$

Then let us calculate the mass of this bit of Earth, with Earth's density $5.5\times {10}^3\text{kg}/{\text{m}}^3$ and assume all the mass is contributed by nucleons of mass $1.67\times {10}^{-27}\text{kg}$.

$\begin{array}{l}=\frac{1.28\times {10}^7{\text{\hspace{0.17em}m}}^3\times 5.5\times {10}^3\text{\hspace{0.17em}kg}/{\text{m}}^3}{1.67\times {10}^{-27}\text{\hspace{0.17em}kg}}\\ =4\times {10}^{37}{\text{\hspace{0.17em}nucleons/m}}^{\text{2}}\end{array}$

Thus, there would be around $4\times {10}^{37}$ nucleons/m².

Simplify the expression

(b)

The probability is the multiple of effective area and number of nucleons can be calculated as:

$\begin{array}{l}=(4\times {10}^{37})\left({10}^{-48}\right)\\ =4\times {10}^{-11}\end{array}$

Thus, the probability is $4\times {10}^{-11}$.