91Ó°ÊÓ

The given data can be listed below as:

• The mass of the lithium nucleus is,$m=5.1×{10}^{-27}\mathrm{kg}$
• The speed of the lithium nucleus is,$v=0.984c$
• The work done by the electric force is,$w=4.7×{10}^{-9}\mathrm{J}$

The energy is neither destroyed, nor it was created; it can only be transferred from one form to another form.

The rest energy is described as the product of the mass of an object and the square of the speed of the light.

The kinetic energy is described as the half of the product of the mass of an object and the square of the velocity of that object.

The equation of the initial particle energy is expressed as:

$E_{particl{e}_{-}\mathrm{i}}=E_{rest}+K$ …(¾±)

Here,$E_{rest}$is the rest energy and Kis the kinetic energy.

The equation of the initial rest energy is expressed as:

$E_{res{t}_{-}\mathrm{i}}=m{c}^2$ …(¾±¾±)

Here,m is the mass of the lithium nucleus andc is the speed of the light.

The equation of the initial kinetic energy is expressed as:

$$\begin{gathered} k_i=\left(y-1\right)m{c}^2 \\ =\left(\frac{1}{\sqrt{1{\displaystyle \frac{v^2}{c^2}}}}-1\right)m{c}^2 \end{gathered}$$ …(¾±¾±¾±)

Here, y is a constant whose value is $\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}},m$is the mass of the lithium nucleus and is the speed of the light.

Substitute the values from equation (ii) and equation (iii) in equation (i).

$$\begin{gathered} E_{particl{e}_{-}\mathrm{i}}=m{c}^2+\left(y-1\right)m{c}^2 \\ =ym{c}^2 \\ =\frac{m{c}^2}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \end{gathered}$$

Substitute the values in the above equation.

$$\begin{gathered} E_{particl{e}_{-}\mathrm{i}}=\frac{\left(5.1×{10}^{-27}\mathrm{kg}\right)\left(3×{10}^{8}m/s\right)}{\sqrt{1-{\displaystyle \frac{{\left(0.984c\right)}^2}{c^2}}}} \\ =\frac{\left(5.1×{10}^{-27}\mathrm{kg}\right)\left(9×{10}^{16}{m}^2/s^2\right)}{\sqrt{1-{\displaystyle \frac{{\left(0.984c\right)}^2}{c^2}}}} \\ =\frac{\left(4.59×{10}^{-10}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{\sqrt{1-\frac{{\left(0.968c\right)}^2}{c^2}}} \\ =\frac{\left(4.59×{10}^{-10}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{\sqrt{0.032}} \end{gathered}$$

Hence, further as:

$$\begin{gathered} E_{paticl{e}_{-}\mathrm{i}}=\frac{\left(4.59×{10}^{-10}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{0178} \\ =2.57×{10}^{-9}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2 \\ =2.57×{10}^{-9}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2×\left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =2.57×{10}^{-9}\mathrm{J} \end{gathered}$$

Thus, the particle energy is $2.57×{10}^{-9}\mathrm{J}$.

Substitute the values in equation (ii).

$$\begin{gathered} E_{res{t}_{-}\mathrm{i}}=\left(5.1×{10}^{-27}\mathrm{kg}\right){\left(3×{10}^{8}m/s\right)}^2 \\ =\left(5.1×{10}^{-27}\mathrm{kg}\right){\left(9×{10}^{16}m/s\right)}^2 \\ =\left(4.59×{10}^{-10}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2×\left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right)\right) \\ =4.59×{10}^{-10}\mathrm{J} \end{gathered}$$

Thus, the value of the rest energy is $4.59×{10}^{-10}\mathrm{J}$.

Substitute the values in equation (iii).

$$\begin{gathered} {\mathrm{k}}_{\mathrm{i}}=\left(\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(0.984\mathrm{c}\right)}^2}{{\mathrm{c}}^2}}}}-1\right)\left(5.1×{10}^{-27}\mathrm{kg}\right){\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)}^2 \\ =\left(\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(0.968\mathrm{c}\right)}^2}{{\mathrm{c}}^2}}}}-1\right)\left(5.1×{10}^{-27}\mathrm{kg}\right)\left(9×{10}^{16}{\mathrm{m}}^2/{\mathrm{s}}^2\right) \\ =\left(\frac{1}{\sqrt{0.032}}-1\right)\left(4.59×{10}^{-10}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\right) \\ =\left(\frac{1}{\sqrt{0.178}}-1\right)\left(4.59×{10}^{-10}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\right) \end{gathered}$$

Hence, further as:

$$\begin{gathered} k_i=\left(5.61-1\right)\left(4.59×{10}^{-10}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\right) \\ =4.61×\left(4.59×{10}^{-10}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\right) \\ =2.1×{10}^{-9}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2×\left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =2.1×{10}^{-9}\mathrm{J} \end{gathered}$$

Thus, the value of the kinetic energy is $2.1×{10}^{-9}$.

The equation of the particle energy after the electric force is expressed as:

$E_f=E_{particl{e}_{-}i}+w$

Here, $E_{particl{e}_{-}i}$is the initial particle energy and w is the work done by the electric force.

Substitute the values in the above equation.

$$\begin{gathered} E_f=2.57×{10}^{-9}\mathrm{J}+4.7×{10}^{-9}\mathrm{J} \\ =7.27×{10}^{-9}\mathrm{J} \end{gathered}$$

Thus, the particle energy after the addition of the electric force is $7.27×{10}^{-9}\mathrm{J}$.

The particle's rest energy after the electric force will be the same as that of the particle's before the electric force because the rest energy is not dependent upon the work done due to the electric force.

Thus, the value of the rest energy is$4.59×{10}^{-10}\mathrm{J}$ .

The equation of the kinetic energy is expressed as:

$k_f=k_i+w$

Here, $k_i$is the initial kinetic energy and w is the work done by the electric force.

Substitute the values in the above equation.

$$\begin{gathered} k_f=2.12×{10}^{-9}\mathrm{J}+4.7×{10}^{-9}\mathrm{J} \\ =6.82×{10}^{-9}\mathrm{J} \end{gathered}$$

Thus, the kinetic energy is$6.82×{10}^{-9}\mathrm{J}$.

The value of the particle energy, rest energy, and kinetic energy in the first case is,$2.57×{10}^{-9}\mathrm{J},459×{10}^{-10}\mathrm{J}\mathrm{and}2.1×{10}^{-9}\mathrm{J}$respectively. The value of the particle energy, rest energy, and kinetic energy in the second case is ,$7.27×{10}^{-9}\mathrm{J},459×{10}^{-10}\mathrm{J}\mathrm{and}6.82×{10}^{-9}\mathrm{J}$respectively.