91Ó°ÊÓ

• Young modulus for aluminum,$\mathrm{Y}=6.2×{10}^{10}\text{ }\mathrm{N}/{\mathrm{m}}^2$
• Mass of one mole aluminum,$\mathrm{m}=27\text{ }\mathrm{g}$
• Density of lead;

$\begin{array}{c}\mathrm{ÒÏ}=11.4\text{ }\mathrm{g}/{\mathrm{cm}}^3\\ =11.4×\frac{{10}^{−3}}{{\left({10}^{−2}\right)}^3}\\ =11400\text{ }\mathrm{kg}/{\mathrm{m}}^3\end{array}$

• Young modulus for aluminum,$\mathrm{Y}=1.6×{10}^{10}\text{ }\mathrm{N}/{\mathrm{m}}^2$
• Mass of one mole aluminum,

$\begin{array}{c}\mathrm{m}=27\text{ }\mathrm{g}\\ =0.207\text{ }\mathrm{kg}\end{array}$

The expression for the young modulus is given by,

$\mathbf{Y}\mathbf{=}\frac{{\mathbf{k}}_{\mathbf{s}}}{\mathbf{d}}$

Here ${\mathbf{k}}_{\mathbf{s}}$is the stiffness of an interatomic bond in a solid, $\mathbf{d}$is the length of an interatomic bond and$\mathbf{Y}$ is the young modulus.

The expression for the angular frequency is given by,

$\mathbf{Ó\neg }\mathbf{=}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}$

Here$\mathbf{Ó\neg }$ is the angular frequency,$\mathrm{k}$ is the spring constant,$\mathrm{m}$ is the mass.

Assume that an aluminum is cube of side $1\mathrm{m}$. Then the number of atoms inside this cube is,

$\begin{array}{c}{\mathrm{N}}_{\mathrm{Al}}=2700\text{ }\mathrm{kg}/{\mathrm{m}}^3\left(\frac{1\text{ }\mathrm{mol}}{0.027\text{ }\mathrm{kg}}\right)\left(\frac{6.02×{10}^{23}\text{ }\mathrm{atoms}}{1\text{ }\mathrm{mol}}\right)\\ =6.02×{10}^{28}\text{ }\mathrm{atoms}\end{array}$

Now, the number of atoms along one edge of the cube is,

$\begin{array}{c}{\mathrm{n}}_{\mathrm{Al}}=\sqrt[3]{6.02×{10}^{28}}\text{ }\mathrm{atoms}\\ =3.92×{10}^9\text{ }\mathrm{atoms}\end{array}$

As the row is $1\mathrm{m}$long, so the diameter of one atom is,

$\begin{array}{c}{\mathrm{d}}_{\mathrm{Al}}=\frac{1\text{ }\mathrm{m}}{3.92×{10}^9\text{ }\mathrm{atoms}}\\ =2.55×{10}^{−10}\text{ }\mathrm{m}\end{array}$

The expression for the stiffness of an interatomic bond in aluminum is,

${\mathrm{k}}_{\mathrm{Al}}={\mathrm{Y}}_{\mathrm{Al}}{\mathrm{d}}_{\mathrm{Al}}$

Substitute$6.2×{10}^{10}\text{ }\mathrm{N}/{\mathrm{m}}^2$for ${\mathrm{Y}}_{\mathrm{Al}}$ and $2.55×{10}^{−10}\text{ }\mathrm{m}$for role="math" localid="1661248912662" ${\mathrm{d}}_{\mathrm{Al}}$into the above equation,

$\begin{array}{c}{\mathrm{k}}_{\mathrm{Al}}=(\mathrm{Y}=6.2×{10}^{10})(2.55×{10}^{−10})\\ =15.8\text{ }\mathrm{N}/\mathrm{m}\end{array}$

Assume that an lead is cube of side $1\text{ }\mathrm{m}$. Then the number of atoms inside this cube is,

$\begin{array}{c}{\mathrm{N}}_{\mathrm{Pb}}=11400\text{ }\mathrm{kg}/{\mathrm{m}}^3\left(\frac{1\text{ }\mathrm{mol}}{0.27\text{ }\mathrm{kg}}\right)\left(\frac{6.02×{10}^{23}\text{ }\mathrm{atoms}}{1\text{ }\mathrm{mol}}\right)\\ =2.54×{10}^{28}\text{ }\mathrm{atoms}\end{array}$

Now, the number of atoms along one edge of the cube is,

$\begin{array}{c}{\mathrm{n}}_{\mathrm{Pb}}=\sqrt[3]{2.54×{10}^{28}}\text{ }\mathrm{atoms}\\ =2.94×{10}^9\text{ }\mathrm{atoms}\end{array}$

As the row is$1\mathrm{m}$long, so the diameter of one atom is,

$\begin{array}{c}{\mathrm{d}}_{\mathrm{Pb}}=\frac{1\text{ }\mathrm{m}}{2.94×{10}^9\text{ }\mathrm{atoms}}\\ =3.4×{10}^{−10}\text{ }\mathrm{m}\end{array}$

The expression for the stiffness of an interatomic bond in aluminum is,

${\mathrm{k}}_{\mathrm{Pb}}={\mathrm{Y}}_{\mathrm{Pb}}{\mathrm{d}}_{\mathrm{Pb}}$

Substitute $6.2×{10}^{10}\text{ }\mathrm{N}/{\mathrm{m}}^2$for ${\mathrm{Y}}_{\mathrm{Pb}}$ and $2.55×{10}^{−10}\text{ }\mathrm{m}$for ${\mathrm{d}}_{\mathrm{Pb}}$into the above equation,

$\begin{array}{c}{\mathrm{k}}_{\mathrm{Pb}}=(\mathrm{Y}=1.6×{10}^{10})(3.4×{10}^{−10})\\ =5.44\text{ }\mathrm{N}/\mathrm{m}\end{array}$

The effective spring stiffness corresponding to the interatomic force for aluminum and lead is,

${\mathrm{k}}_{\mathrm{eff}}={\mathrm{k}}_{\mathrm{Al}}+{\mathrm{k}}_{\mathrm{Pb}}$

Substitute $15.8\text{ }\mathrm{N}/\mathrm{m}$for ${\mathrm{k}}_{\mathrm{Al}}$and $5.44\text{ }\mathrm{N}/\mathrm{m}$for ${\mathrm{k}}_{\mathrm{Pb}}$into the above equation,

$\begin{array}{c}{\mathrm{k}}_{\mathrm{eff}}=15.8+5.44\\ =21.24\text{ }\mathrm{N}/\mathrm{m}\end{array}$

The expression for the angular frequency of ${\mathrm{H}}_2$molecule is given by,

${\mathrm{Ó\neg }}_{{\mathrm{H}}_2}=\sqrt{\frac{{\mathrm{k}}_{{\mathrm{H}}_2}}{{\mathrm{m}}_{{\mathrm{H}}_2}}}$

The corresponding oscillating frequency of ${\mathrm{H}}_2$molecule is given by,

${\mathrm{f}}_{{\mathrm{H}}_2}=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{{\mathrm{k}}_{{\mathrm{H}}_2}}{{\mathrm{m}}_{{\mathrm{H}}_2}}}$

Substitute$21.24\text{ }\mathrm{N}/\mathrm{m}$ for ${\mathrm{k}}_{{\mathrm{H}}_2}$and$2(1.67×{10}^{−27}\text{ }\mathrm{kg})$ for ${\mathrm{m}}_{{\mathrm{H}}_2}$into the above equation,

$\begin{array}{c}{\mathrm{f}}_{{\mathrm{H}}_2}=\frac{1}{2\mathrm{π}}\sqrt{\frac{21.24}{2(1.67×{10}^{−27})}}\\ =1.3×{10}^{13}\text{ }\mathrm{Hz}\end{array}$

Therefore, the approximate frequency of hydrogen molecule is $1.3×{10}^{13}\text{ }\mathrm{Hz}$.

The corresponding oscillating frequency of $O_2$molecule is given by,

${\mathrm{f}}_{{\mathrm{O}}_2}=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{{\mathrm{k}}_{{\mathrm{O}}_2}}{{\mathrm{m}}_{{\mathrm{O}}_2}}}$

Substitute$21.24\text{ }\mathrm{N}/\mathrm{m}$ for ${\mathrm{k}}_{{\mathrm{O}}_2}$and $32(1.67×{10}^{−27}\text{ }\mathrm{kg})$for ${\mathrm{m}}_{{\mathrm{O}}_2}$into the above equation,

$\begin{array}{c}{\mathrm{f}}_{{\mathrm{O}}_2}=\frac{1}{2\mathrm{π}}\sqrt{\frac{21.24}{32(1.67×{10}^{−27})}}\\ =3.2×{10}^{12}\text{ }\mathrm{Hz}\end{array}$

Therefore, the approximate frequency of oxygen molecule is $3.2×{10}^{12}\text{ }\mathrm{Hz}$.

The corresponding oscillating frequency of ${\mathrm{D}}_2$molecule is given by,

${\mathrm{f}}_{{\mathrm{D}}_2}=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{{\mathrm{k}}_{{\mathrm{D}}_2}}{{\mathrm{m}}_{{\mathrm{D}}_2}}}$

Substitute $21.24\text{ }\mathrm{N}/\mathrm{m}$ for ${\mathrm{k}}_{{\mathrm{D}}_2}$and $4(1.67×{10}^{−27}\text{ }\mathrm{kg})$ for${\mathrm{m}}_{{\mathrm{D}}_2}$ into the above equation,

$\begin{array}{c}{\mathrm{f}}_{{\mathrm{O}}_2}=\frac{1}{2\mathrm{π}}\sqrt{\frac{21.24}{4(1.67×{10}^{−27})}}\\ =8.97×{10}^{12}\text{ }\mathrm{Hz}\end{array}$

Therefore, the approximate frequency of deuterium molecule is $8.9×{10}^{12}\text{ }\mathrm{Hz}$.

From the expression

$\mathrm{Ó\neg }=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

Angular frequency is directly proportional to the square root of the spring stiffness and inversely proportional to the square root of the mass of the molecule.

As in the above cases spring stiffness remains constant, the atomic mass of the hydrogen and deuterium is not changing, so the ratio of the frequencies of these given molecule is quite accurate.