91影视

Cross-sectional area $=1{\text{鈥尘}}^2$.

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.

(a)

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

$\begin{array}{l}{\mathrm{蚁}}_{\mathrm{E}}=\frac{(5.98脳{10}^{24}\text{鈥塳驳})}{\frac{4}{3}\mathrm{蟺}{(6.37脳{10}^6\text{鈥尘})}^3}\\ {\mathrm{蚁}}_{\mathrm{E}}=5.5脳{10}^3{\text{鈥塳驳/m}}^{\text{3}}\end{array}$

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

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

$\begin{array}{l}\left(2{\mathrm{R}}_{\mathrm{E}}\right)(1{\text{鈥尘}}^3)=2(6.37脳{10}^6\text{鈥尘})\left(1{\text{鈥尘}}^2\right)\\ \left(2{\mathrm{R}}_{\mathrm{E}}\right)(1\text{鈥�}{\text{m}}^3)=1.27脳{10}^7{\text{鈥尘}}^3\end{array}$

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

$\begin{array}{l}=\frac{1.28脳{10}^7{\text{鈥尘}}^3脳5.5脳{10}^3\text{鈥塳驳}/{\text{m}}^3}{1.67脳{10}^{鈭�27}\text{鈥塳驳}}\\ =4脳{10}^{37}{\text{鈥塶耻肠濒别辞苍蝉/尘}}^{\text{2}}\end{array}$

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

(b)

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

$\begin{array}{l}=(4脳{10}^{37})\left({10}^{鈭�48}\right)\\ =4脳{10}^{鈭�11}\end{array}$

Thus, the probability is $4脳{10}^{鈭�11}$.