91影视

The Hypothetical nucleus is ${}_{119}^{288}\text{X}$.

Semi-empirical binding energy is given by the expression: $\mathbf{BE}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{A}\mathbf{鈭�}{\mathbf{c}}_{\mathbf{2}}{\mathbf{A}}^{\mathbf{2}\mathbf{/}\mathbf{3}}\mathbf{鈭�}{\mathbf{c}}_{\mathbf{3}}\frac{\mathbf{Z}\mathbf{(}\mathbf{Z}\mathbf{鈭�}\mathbf{1}\mathbf{)}}{{\mathbf{A}}^{\mathbf{1}\mathbf{/}\mathbf{3}}}\mathbf{鈭�}{\mathbf{c}}_{\mathbf{4}}\frac{{\mathbf{(}\mathbf{N}\mathbf{鈭�}\mathbf{Z}\mathbf{)}}^{\mathbf{2}}}{\mathbf{A}}$ (1)

Where,${\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{15}\mathbf{.8}\mathbf{\text{鈥�}}\mathbf{MeV}\mathbf{,}\mathbf{\text{鈥�}}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{17}\mathbf{.8}\mathbf{\text{鈥�}}\mathbf{MeV}\mathbf{,}\mathbf{\text{鈥�}}{\mathbf{c}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.71}\mathbf{\text{鈥�}}\mathbf{MeV}\mathbf{,}\mathbf{\text{鈥嬧赌�}}{\mathbf{c}}_{\mathbf{4}}\mathbf{=}\mathbf{23}\mathbf{.7}\mathbf{\text{鈥�}}\mathbf{MeV}$,$\mathbf{A}$is the mass number,$\mathbf{Z}$is the atomic number, and$\mathbf{N}$is the number of neutrons.

(a)
$\begin{array}{l}{\text{Foratom}}_{119}^{288}\text{X}\\ \mathrm{A}=288,\text{鈥�}\mathrm{Z}=119,\text{鈥�}\mathrm{N}=\mathrm{A}鈭�\mathrm{Z}=288鈭�119=169\end{array}$

Substitute values in equation (1), the binding energy of ${}_{119}^{288}\mathrm{X}$ is:

$\begin{array}{c}\text{BE}=\left(15.8\text{鈥塎别痴}\right)\left(288\right)鈭�\left(17.8\text{鈥塎别痴}\right){\left(288\right)}^{2/3}鈭�\left(0.71\text{鈥塎别痴}\right)\frac{\left(119\right)(119鈭�1)}{{\left(288\right)}^{1/3}}\\ 鈭�\left(23.7\text{鈥塎别痴}\right)\frac{{(169鈭�119)}^2}{288}\\ =2058.70\text{鈥塎别痴}\end{array}$

Suppose it underwent alpha decay, then it will turn into,

${}_{119}^{288}\mathrm{X}鈫�{}_2{}^4\mathrm{伪}{+}_{117}^{284}\mathrm{Y}$

For alpha particle,

$\mathrm{A}=4,\text{鈥�}\mathrm{Z}=2,\text{鈥�}\mathrm{N}=4鈭�2=2$

Substitute values in equation (1), the binding energy of the alpha particle is:

$\begin{array}{c}\text{BE}=\left(15.8\text{鈥塎别痴}\right)\left(4\right)鈭�(17.8\text{鈥塎别痴)}{\left(4\right)}^{2/3}鈭�(0.71\text{鈥塎别痴})\frac{\left(2\right)(2鈭�1)}{{\left(4\right)}^{1/3}}鈭�(23.7\text{鈥塎别痴})\frac{{(2鈭�2)}^2}{4}\\ =17.45\text{鈥塎别痴}\end{array}$

The binding energy of the other particle is,

$\mathrm{A}=284,\text{鈥�}\mathrm{Z}=117,\text{鈥�}\mathrm{N}=284鈭�117=167$

Substitute values in equation (1), as:

$\begin{array}{c}\text{BE}=\left(15.8\text{鈥塎别痴}\right)\left(284\right)鈭�\left(17.8\text{鈥塎别痴}\right){\left(284\right)}^{2/3}鈭�\left(0.71\text{鈥塎别痴}\right)\frac{\left(117\right)(117鈭�1)}{{\left(284\right)}^{1/3}}\\ 鈭�\left(23.7\text{鈥塎别痴}\right)\frac{{(167鈭�117)}^2}{284}\\ =2043.52\text{鈥塎别痴}\end{array}$

Since the binding energy difference is:

$\begin{array}{l}=2058.70\text{鈥塎别痴}鈭�17.45\text{鈥塎别痴}鈭�2043.52\text{鈥塎别痴}\\ =鈭�2.27\text{鈥塎别痴}\end{array}$.

This decay process is possible because the binding energy increases, making the particle more stable.

Thus,$\mathrm{伪}鈭�$ Decay is possible.

(b)

Suppose it underwent beta-plus decay. It will turn into:

${}_{119}^{288}\text{X}{鈫�}_1^0\mathrm{e}{+}_{118}^{288}\text{Y}$.

The binding energy of the other particle can be calculated as:

$\mathrm{A}=288,\text{鈥�}\mathrm{Z}=118,\text{鈥�}\mathrm{N}=288鈭�118=170$

Substitute values in equation (1), as:

$\begin{array}{c}\text{BE}=\left(15.8\text{鈥塎别痴}\right)\left(288\right)鈭�\left(17.8\text{鈥塎别痴}\right){\left(288\right)}^{2/3}鈭�\left(0.71\text{鈥塎别痴}\right)\frac{\left(118\right)(118鈭�1)}{{\left(288\right)}^{1/3}}\\ 鈭�\left(23.7\text{鈥塎别痴}\right)\frac{{(170鈭�118)}^2}{288}\\ =2067.28\text{鈥塎别痴}\end{array}$

Since this particle's binding energy is higher than the original particle by several MeVs, and thus more stable, this process is possible.

Thus, ${\mathrm{尾}}^{+}鈭�$Decay is possible.

(c)

Suppose it underwent beta-minus decay. It would turn into,

${}_{119}^{288}\text{X}{鈫�}_{鈭�1}^0\mathrm{e}{+}_{120}^{288}\text{Y}$

The binding energy of the other particle can be calculated as:

$\mathrm{A}=288,\text{鈥�}\mathrm{Z}=120,\text{鈥�}\mathrm{N}=288鈭�120=168$

Substitute values in equation (1), as:

$\begin{array}{c}\text{BE}=\left(15.8\text{鈥塎别痴}\right)\left(288\right)鈭�\left(17.8\text{鈥塎别痴}\right){\left(288\right)}^{2/3}鈭�\left(0.71\text{鈥塎别痴}\right)\frac{\left(120\right)(120鈭�1)}{{\left(288\right)}^{1/3}}\\ 鈭�\left(23.7\text{鈥塎别痴}\right)\frac{{(168鈭�120)}^2}{288}\\ =2049.24\text{鈥塎别痴}\end{array}$

Since this binding energy is smaller than the original particle, this process is impossible because the original particle is more stable

.

${\mathrm{尾}}^{鈭�}鈭�$Decay is not possible.