Chapter 30, Problem 33P

To determine

Number of electrons and quark particle in $1.00\text{L}$.

Answer to Problem 33P

There are $3.34\times {10}^{26}$ electrons, $9.36\times {10}^{26}$ up quarks and $8.70\times {10}^{26}$ down quarks.

Explanation of Solution

Given info:

Mass of water is $1.00\text{L}$.

Avogadro number is $6.02\times {10}^{23}\text{molecules/mol}$.

The number of gram per mole of water is $18.0\text{g/mol}$.

Formula to the total number of molecules in water is,

$N=\frac{m}{n}{N}_A$

•   $N$ is the number of molecules
• $m$ is the mass of water
• $n$ is the number of gram per mole of water
• $N_A$ is the Avogadro number

Substitute $1.00\text{L}$ for $m$, $6.02\times {10}^{23}\text{molecules/mol}$ for $N_A$ and $18.0\text{g/mol}$ for $n$ to find $N$.

$\begin{array}{c}N=\frac{\left(1.00\text{L}\right)\left(\frac{1000\text{g}}{1\text{L}}\right)}{\left(18.0\text{g/mol}\right)}\left(6.02\times {10}^{23}\text{molecules/mol}\right)\\ =3.34\times {10}^{25}\text{molecules}\end{array}$

Each molecule contains $10$ protons, $10$ electrons, and $8$ neutrons.

Formula to the total number of electron in $1.00\text{L}$ of water is,

$N_e=n_{e}N$

•   $N$ is the number of molecules
• $n_e$ is the number of electron in one molecule of water
• $N_e$ is the total number of electron in $1.00\text{L}$ of water

Substitute $3.34\times {10}^{25}\text{molecules}$ for $N$ and $10\text{electrons/molecule}$ for $n_e$ to find $N_e$.

$\begin{array}{c}{N}_e=\left(10\text{electrons/molecule}\right)\left(3.34\times {10}^{25}\text{molecules}\right)\\ =3.34\times {10}^{26}\text{electrons}\end{array}$

Thus, the total number of electrons in $1.00\text{L}$ of water is $3.34\times {10}^{26}\text{electrons}$.

Formula to the total number of proton in $1.00\text{L}$ of water is,

$N_p=n_{p}N$

• $n_p$ is the number of proton in one molecule of water
• $N_p$ is the total number of proton in $1.00\text{L}$ of water

Substitute $3.34\times {10}^{25}\text{molecules}$ for $N$ and $10\text{protons/molecule}$ for $n_p$ to find $N_p$.

$\begin{array}{c}{N}_p=\left(10\text{protons/molecule}\right)\left(3.34\times {10}^{25}\text{molecules}\right)\\ =3.34\times {10}^{26}\text{protons}\end{array}$

Thus, the total number of protons in $1.00\text{L}$ of water is $3.34\times {10}^{26}\text{protons}$.

Formula to the total number of neutron in $1.00\text{L}$ of water is,

$N_n=n_{n}N$

• $n_n$ is the number of neutron in one molecule of water
• $N_n$ is the total number of neutron in $1.00\text{L}$ of water

Substitute $3.34\times {10}^{25}\text{molecules}$ for $N$ and $8\text{neutron/molecule}$ for $n_n$ to find $N_n$.

$\begin{array}{c}{N}_n=\left(8\text{neutron/molecule}\right)\left(3.34\times {10}^{25}\text{molecules}\right)\\ =2.68\times {10}^{26}\text{protons}\end{array}$

Each proton contains $2$ up quarks and $1$ down quark. Each neutron contains $1$ up quark and $2$ down quarks.

Formula to the total number of up quarks in $1.00\text{L}$ of water is,

$N_u=2N_p+N_n$

•   $N_u$ is the total number of up quarks

Substitute $3.34\times {10}^{26}\text{protons}$ for $N_p$ and $2.26\times {10}^{26}\text{neutrons}$ for $N_n$ to find $N_u$.

$\begin{array}{c}{N}_u=2\left(3.34\times {10}^{26}\text{protons}\right)+\left(2.26\times {10}^{26}\text{neutrons}\right)\\ =9.36\times {10}^{26}\text{up quarks}\end{array}$

Formula to the total number of down quarks in $1.00\text{L}$ of water is,

$N_d=N_p+2N_n$

•   $N_d$ is the total number of down quarks

Substitute $3.34\times {10}^{26}\text{protons}$ for $N_p$ and $2.26\times {10}^{26}\text{neutrons}$ for $N_n$ to find $N_d$.

$\begin{array}{c}{N}_d=\left(3.34\times {10}^{26}\text{protons}\right)+2\left(2.26\times {10}^{26}\text{neutrons}\right)\\ =8.70\times {10}^{26}\text{up quarks}\end{array}$

Thus, the total number of electrons in $1.00\text{L}$ of water is $3.34\times {10}^{26}\text{electrons}$.

Conclusion:

There are $3.34\times {10}^{26}$ electrons, $9.36\times {10}^{26}$ up quarks and $8.70\times {10}^{26}$ down quarks.