91Ó°ÊÓ

A fairly low-mass nuclide ${}^{55}\mathrm{Mn}$and a fairly high-mass nuclide ${}^{209}\mathrm{Bi}$are given.

The mass density of an atom as follows:

${\mathbf{p}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{N}}{{\mathbf{N}}_{\mathbf{A}}\mathbf{V}}$ ….. (i)

Here, M is the molar mass of the substance and.

The charge density of an atom is as follows:

${\mathbf{p}}_{\mathbf{q}}\mathbf{=}\frac{\mathbf{Ze}}{\mathbf{V}}$ …… (ii)

The volume of a spherical body is as follows:

$\mathbf{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{Ï„Ï„°ù}}^{\mathbf{3}}$ …… (iii)

The radius of a nucleus is as follows:

$\mathbf{r}\mathbf{=}{\mathbf{r}}_{\mathbf{0}}{\mathbf{A}}^{\frac{\mathbf{1}}{\mathbf{3}}}$ …… (iv)

Here, A is the atomic mass of the substance and ${\mathrm{r}}_0=1.2×{10}^{-15}\mathrm{m}$.

Atomic mass of the low-mass nuclide ${}^{55}\mathrm{Mn}$,A = 55

Molar mass of,${}^{55}\mathrm{Mn},\mathrm{M}=55\frac{\mathrm{g}}{\mathrm{mol}}\mathrm{or}0.055\frac{\mathrm{kg}}{\mathrm{mol}}$

Using the given data and equations (iii) and (iv) in equation (i), Determine the value of the nuclear mass density of the low-mass nuclide as follows:

$$\begin{gathered} {\mathrm{p}}_{\mathrm{m}}=\frac{0.055{}^{55}\frac{\mathrm{kg}}{\mathrm{mol}}}{\left({\displaystyle \frac{4\mathrm{π}}{3}}\right){\left[\left(1.2×{10}^{-15}\mathrm{m}\right){\left(55\right)}^{{\displaystyle \frac{1}{3}}}\right]}^3\left(6.022×{10}^{23}\mathrm{mol}\right)} \\ =2.3×{10}^{17}\frac{\mathrm{kg}}{{\mathrm{m}}^3} \end{gathered}$$

Hence, the value of the density is $2.3×{10}^{17}\frac{\mathrm{kg}}{{\mathrm{m}}^3}$.

Atomic mass of the high-mass nuclide,${}^{209}\mathrm{Bi},\mathrm{A}=209$

Molar mass of,${}^{209}\mathrm{Bi},\mathrm{A}=209\frac{\mathrm{g}}{\mathrm{mol}}\mathrm{or}0.209\frac{\mathrm{kg}}{\mathrm{mol}}$

Using the given data and equations (iii) and (iv) in equation (i), determine the value of the nuclear mass density of the low-mass nuclide as follows:

$$\begin{gathered} {\mathrm{p}}_{\mathrm{m}}=\frac{0.209\frac{\mathrm{kg}}{\mathrm{mol}}}{\left({\displaystyle \frac{4\mathrm{π}}{3}}\right){\left[\left(1.2×{10}^{-15}\mathrm{m}\right){\left(209\right)}^{{\displaystyle \frac{1}{3}}}\right]}^3\left(6.022×{10}^{23}\mathrm{mol}\right)} \\ =2.3×{10}^{17}\frac{\mathrm{kg}}{{\mathrm{m}}^3} \end{gathered}$$

Hence, the value of the density is localid="1661926851336" $2.3×{10}^{17}\frac{\mathrm{kg}}{{\mathrm{m}}^3}$.

By substituting equation (iv) in equation (iii) solve as:

$$\begin{gathered} V\mathrm{Î\pm }{r}^3 \\ V\mathrm{Î\pm }\left(r_0{A}^{1/3}\right) \\ V\mathrm{Î\pm }A \end{gathered}$$

Again, using the above value and equation (i):

$$\begin{gathered} p_m\mathrm{Î\pm }\frac{A}{V} \\ {p}_m\mathrm{Î\pm }\frac{A}{A} \\ {p}_m\mathrm{Î\pm }\mathrm{constant} \end{gathered}$$

Hence, the nuclear mass density is constant for all the nuclides.

Charge of the low-mass nuclide,${}^{55}\mathrm{Mn},\mathrm{Ze}=25\mathrm{e}$

Molar mass of,${}^{55}\mathrm{Mn},\mathrm{M}=55\frac{\mathrm{g}}{\mathrm{mol}}\mathrm{or}0.055\frac{\mathrm{kg}}{\mathrm{mol}}$

Using the given data and equations (iii) and (iv) in equation (ii), we can get the value of the nuclear charge density of the low-mass nuclide as follows:

$$\begin{gathered} {\mathrm{p}}_{\mathrm{q}}=\frac{25\left(1.6×{10}^{-19}\mathrm{C}\right)}{\left({\displaystyle \frac{4\mathrm{π}}{3}}\right){\left[\left(1.2×{10}^{-15}\mathrm{m}\right){\left(55\right)}^{{\displaystyle \frac{1}{3}}}\right]}^3} \\ =1.0×{10}^{25}\frac{\mathrm{c}}{{\mathrm{m}}^3} \end{gathered}$$

Hence, the value of the density is$1.0×{10}^{25}\frac{\mathrm{c}}{{\mathrm{m}}^3}$ .

Charge of the high-mass nuclide,${}^{209}\mathrm{Bi},\mathrm{Ze}=83\mathrm{e}$

Molar mass of,${}^{209}\mathrm{Bi},\mathrm{M}=209\frac{\mathrm{g}}{\mathrm{mol}}\mathrm{or}0.209\frac{\mathrm{kg}}{\mathrm{mol}}$

Using the given data and equations (iii) and (iv) in equation (ii), determine the value of the nuclear charge density of the low-mass nuclide as follows:

$$\begin{gathered} {\mathrm{p}}_{\mathrm{q}}=\frac{83\left(1.6×{10}^{-19}\mathrm{C}\right)}{\left({\displaystyle \frac{4\mathrm{π}}{3}}\right){\left[\left(1.2×{10}^{-15}\mathrm{m}\right){\left(209\right)}^{{\displaystyle \frac{1}{3}}}\right]}^3} \\ =8.8×{10}^{24}\frac{\mathrm{c}}{{\mathrm{m}}^3} \end{gathered}$$

Hence, the value of the density is$8.8×{10}^{24}\frac{\mathrm{c}}{{\mathrm{m}}^3}$ .

By substituting equation (iv) in equation (iii), we can get that

$$\begin{gathered} V\mathrm{Î\pm }{r}^3 \\ V\mathrm{Î\pm }\left(r_0{A}^{\frac{1}{3}}\right) \\ V\mathrm{Î\pm }A \end{gathered}$$

Again, using the above value and equation (ii), we observe that

$$\begin{gathered} p_q\mathrm{Î\pm }\frac{Z}{V} \\ {p}_q\mathrm{Î\pm }\frac{Z}{A} \end{gathered}$$

The charge density$p_q$ should gradually decrease, since for large nuclides.

Hence, the nuclear charge density should decrease with increase in atomic mass.