Question

A lithium nucleus has mass$\mathbf{5}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{}\mathbf{kg}$If its speed is$\mathbf{0}\mathbf{.}\mathbf{984}\mathbf{c}$(that is,$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{/}\mathbf{c}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{984}$, what are the values of the particle energy, the rest energy, and the kinetic energy?

Next an electric force acts on the lithium nucleus and does$\mathbf{4}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}$ of work on the particle. Now what are the values of the particle energy, the rest energy, and the kinetic energy?

Short answer

The value of the particle energy, rest energy, and kinetic energy in the first case are,$2.57\times {10}^{-9}\mathrm{J},4.59\times {10}^{-10}\mathrm{J}\mathrm{and}2.1\times {10}^{-9}\mathrm{J}$ respectively. The value of the particle energy, rest energy, and the kinetic energy in the second case are $7.27\times {10}^{-9}\mathrm{J},4.59\times {10}^{-10}\mathrm{J}\mathrm{and}6.82\times {10}^{-9}\mathrm{J}$ respectively.

Step-by-step solution

Identification of given data

The given data can be listed below as:

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

Significance of the energy principle, rest energy, and kinetic energy

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.

Determination of the particle energy in the first case

The equation of the initial particle energy is expressed as:

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

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

The equation of the initial rest energy is expressed as:

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

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:

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

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.

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

Hence, further as:

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

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

Determination of the rest energy in the first case

Substitute the values in equation (ii).

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

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

Determination of the kinetic energy in the first case

Substitute the values in equation (iii).

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

Hence, further as:

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

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

Determination of the particle energy in the second case

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.

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

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

Determination of the rest energy in the second case

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$4.59\times {10}^{-10}\mathrm{J}$ .

Determination of the kinetic energy in the second case

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.

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

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

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