Chapter 31, Problem 32P

(a)

To determine

The number of electrons and the number of each species of quark in 1L water.

Answer to Problem 32P

The number of electrons is $\underset{\_}{3.34\times {10}^{26}\text{electrons}}$.

The numbers of each species of quark in 1L water are $9.36\times {10}^{26}\text{ up quarks}$ and $8.70\times {10}^{26}\text{ down quarks}$.

Explanation of Solution

Write the expression for the number of protons.

$N_p=V\left(\frac{N_A}{m}\right)\left(n_p\right)$        (I)

Here, $N_p$ is the number of proton, $V$ is the volume, $N_A$ is the Avogadro number, $m$ is the mass and $n_p$ is the number of proton per molecule.

Write the expression for the number of neutrons.

$N_n=V\left(\frac{N_A}{m}\right)\left(n_n\right)$        (II)

Here, $N_n$ is the number of neutron and $n_n$ is the number of neutron per molecule.

Write the quark configuration of proton.

$p=uud$

Write the quark configuration of neutron.

$n=udd$

Write the expression for the number of up quark.

$N_{\text{up quark}}=2N_p+N_n$        (III)

Here, $N_{\text{up quark}}$ is the number of up quark.

Write the expression for the number of up quark.

$N_{\text{down quark}}=2N_n+N_p$        (IV)

Here, $N_{\text{down quark}}$ is the number of down quark.

Conclusion:

To get neutral atom, the number proton and electron will be same.

Substitute, $1\text{L}$ for $V$, $6.02\times {10}^{23}\text{molecules}$ for $N_A$, $18.0\text{g}$ for $m$, $10\text{proton/molecule}$ for $n_p$ in equation (I) to find $N_p$.

$\begin{array}{c}{N}_p=\left[\left(1\text{L}\right)\left(\frac{1000\text{g}}{1\text{L}}\right)\right]\left(\frac{6.02\times {10}^{23}\text{molecules}}{18.0\text{g}}\right)\left(10\text{proton/molecule}\right)\\ =3.34\times {10}^{26}\text{ protons}\end{array}$

Thus, the number of electrons is $\underset{\_}{3.34\times {10}^{26}\text{electrons}}$.

Substitute, $1\text{L}$ for $V$, $6.02\times {10}^{23}\text{molecules}$ for $N_A$, $18.0\text{g}$ for $m$, $8\text{proton/molecule}$ for $n_n$ in equation (I) to find $N_n$.

$\begin{array}{c}{N}_p=\left[\left(1\text{L}\right)\left(\frac{1000\text{g}}{1\text{L}}\right)\right]\left(\frac{6.02\times {10}^{23}\text{molecules}}{18.0\text{g}}\right)\left(8\text{proton/molecule}\right)\\ =2.68\times {10}^{26}\text{ neutrons}\end{array}$

Substitute, $3.34\times {10}^{26}$ for $N_p$, $2.68\times {10}^{26}$ for $N_n$ in equation (III) to find $N_{\text{up quark}}$.

$\begin{array}{c}{N}_{\text{up quark}}=2\left(3.34\times {10}^{26}\right)+\left(2.68\times {10}^{26}\right)\\ =9.36\times {10}^{26}\text{ up quarks}\end{array}$

Substitute, $3.34\times {10}^{26}$ for $N_p$, $2.68\times {10}^{26}$ for $N_n$ in equation (IV) to find $N_{\text{down quark}}$.

$\begin{array}{c}{N}_{\text{down quark}}=2\left(2.68\times {10}^{26}\right)+\left(3.34\times {10}^{26}\right)\\ =8.70\times {10}^{26}\text{ down quarks}\end{array}$

Thus, the numbers of each species of quark in 1L water are $9.36\times {10}^{26}\text{ up quarks}$ and $8.70\times {10}^{26}\text{ down quarks}$.

(b)

To determine

The number of electrons and the number of each species of quark in body water.

Answer to Problem 32P

The number of electrons is $\underset{\_}{~{10}^{28}\text{electrons}}$.

The numbers of each species of quark in body water are $~{10}^{29}\text{ up quarks}$ and $~{10}^{29}\text{ down quarks}$.

Explanation of Solution

In human body total amount of water is 65kg.

Write the expression for the number of electron in human body.

$N_{e\left(\text{Human}\right)}=V_{\text{human}}{N}_p$        (V)

Here, $N_{e\left(\text{human}\right)}$ is the number of electron in human body, $V_{\text{human}}$ is the total volume of water in human body.

Write the expression for the number of up quark in human body.

$N_{\text{up quark}\left(\text{human}\right)}=V_{\text{human}}{N}_{\text{up quark}}$        (VI)

Here, $N_{\text{up quark}\left(\text{human}\right)}$ is the number of up quark in human body.

Write the quark configuration of down quark in human body.

$N_{\text{down quark}\left(\text{human}\right)}=V_{\text{human}}{N}_{\text{down quark}}$        (VII)

Here, $N_{\text{down quark}\left(\text{human}\right)}$ is the number of down quark in human body.

Conclusion:

Substitute, $65\text{kg}$ for $V_{\text{human}}$, $3.34\times {10}^{26}\text{ protons}$ for $N_p$ in equation (V) to find $N_{e\left(\text{human}\right)}$.

$\begin{array}{c}{N}_{e\left(\text{Human}\right)}=\left(65\text{kg}\right)\left(3.34\times {10}^{26}\text{ protons}\right)\\ ~{10}^{28}\text{electrons}\end{array}$

Thus, the number of electrons is $\underset{\_}{~{10}^{28}\text{electrons}}$.

Substitute, $65\text{kg}$ for $V_{\text{human}}$, $9.36\times {10}^{26}\text{ up quarks}$ for $N_{\text{up quark}}$ in equation (VI) to find $N_{\text{up quark}\left(\text{human}\right)}$.

$\begin{array}{c}{N}_{\text{up quark}\left(\text{human}\right)}=\left(65\text{kg}\right)\left(9.36\times {10}^{26}\text{ up quarks}\right)\\ ~{10}^{29}\text{ up quarks}\end{array}$

Substitute, $65\text{kg}$ for $V_{\text{human}}$, $8.70\times {10}^{26}\text{ down quarks}$ for $N_{\text{down quark}}$ in equation (VII) to find $N_{\text{down quark}\left(\text{human}\right)}$.

$\begin{array}{c}{N}_{\text{down quark}\left(\text{human}\right)}=\left(65\text{kg}\right)\left(8.70\times {10}^{26}\text{ down quarks}\right)\\ ~{10}^{29}\text{ down quarks}\end{array}$

Thus, the numbers of each species of quark in body water are $~{10}^{29}\text{ up quarks}$ and $~{10}^{29}\text{ down quarks}$.