Chapter 16, Problem 19P

A proton is located at the origin, and a second proton is located on the x-axis at x = 6.00 fm (1 fm = 10^-15 m). (a) Calculate the electric potential energy associated with this configuration. (b) An alpha particle (charge = 2e, mass = 6.64 × 10⁻²⁷ kg) is now placed at (x, y) = (3.00, 3.00) fm. Calculate the electric potential energy associated with this configuration. (c) Starting with the three-particle system, find the change in electric potential energy if the alpha particle is allowed to escape to infinity while the two protons remain fixed in place. (Throughout, neglect any radiation effects.) (d) Use conservation of energy to calculate the speed of the alpha particle at infinity. (e) If the two protons are released from rest and the alpha panicle remains fixed, calculate the speed of the protons at infinity.

To determine

The electric potential energy.

Answer to Problem 19P

The electric potential energy is $3.84\times {10}^{-14}\text{J}$.

Explanation of Solution

Given info: Proton is located at x = 6.00 fm.

Explanation:

Formula to calculate the electric potential energy is,

$P{E}_1=\frac{k_e{e}^2}{x}$

• $k_e$ is the coulomb constant.
• e is the elementary charge.

Substitute 6.00 fm for x, $\begin{array}{l}1.6\times {10}^{-19}\text{C}\hfill \end{array}$ for e and $\begin{array}{l}8.99\times {10}^9 {\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\hfill \end{array}$ for $k_e$

$\begin{array}{c}P{E}_1=\frac{\left(\begin{array}{l}8.99\times {10}^9 {\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\hfill \end{array}\right){\left(\begin{array}{l}1.6\times {10}^{-19}\text{C}\hfill \end{array}\right)}^2}{6.00\text{fm}}\\ =\frac{\left(\begin{array}{l}8.99\times {10}^9 {\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\hfill \end{array}\right){\left(\begin{array}{l}1.6\times {10}^{-19}\text{C}\hfill \end{array}\right)}^2}{6.00\times {10}^{-15}\text{m}}\\ =3.84\times {10}^{-14}\text{J}\end{array}$

Conclusion:

The electric potential energy is $3.84\times {10}^{-14}\text{J}$.

To determine

The new electric potential energy.

Answer to Problem 19P

The new electric potential energy is $2.55\times {10}^{-13}\text{J}$.

Explanation of Solution

Given info: Proton is located at x = 6.00 fm.

Explanation:

Formula to calculate the new electric potential energy is,

$P{E}_2=P{E}_1+2\left[\frac{k_e\left(e\right)\left(2e\right)}{r}\right]$

• $k_e$ is the coulomb constant.
• e is the elementary charge.
• r is the distance of the alpha particle from the origin.

The distance of the alpha particle from the origin is,

$\begin{array}{c}r=\sqrt{{\left(3.00\text{fm}\right)}^2+{\left(3.00\text{fm}\right)}^2}\\ =4.243\text{fm}\end{array}$

Substitute $3.84\times {10}^{-14}\text{J}$ for $P{E}_1$, 4.243 fm for r, $\begin{array}{l}1.6\times {10}^{-19}\text{C}\hfill \end{array}$ for e and $\begin{array}{l}8.99\times {10}^9 {\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\hfill \end{array}$ for $k_e$

$\begin{array}{c}P{E}_2=\left(3.84\times {10}^{-14}\text{J}\right)+2\left[\frac{\left(\begin{array}{l}8.99\times {10}^9 {\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\hfill \end{array}\right){\left(\begin{array}{l}1.6\times {10}^{-19}\text{C}\hfill \end{array}\right)}^2}{4.243\text{fm}}\right]\\ =2.55\times {10}^{-13}\text{J}\end{array}$

Conclusion:

The new electric potential energy is $2.55\times {10}^{-13}\text{J}$.

To determine

The change in electric potential energy.

Answer to Problem 19P

The change in electric potential energy is $-2.17\times {10}^{-13}\text{J}$.

Explanation of Solution

Given info: Proton is located at x = 6.00 fm.

Explanation:

Formula to calculate the change in electric potential energy is,

$\Delta PE=P{E}_1-P{E}_2$

Substitute $3.84\times {10}^{-14}\text{J}$ for $P{E}_1$ and $2.55\times {10}^{-13}\text{J}$ for $P{E}_2$

$\begin{array}{c}\Delta PE=\left(3.84\times {10}^{-14}\text{J}\right)-\left(2.55\times {10}^{-13}\text{J}\right)\\ =-2.17\times {10}^{-13}\text{J}\end{array}$

Conclusion:

The change in electric potential energy is $-2.17\times {10}^{-13}\text{J}$.

To determine

The speed of alpha particle.

Answer to Problem 19P

The speed of alpha particle is $8.08\times {10}^6\text{m/s}$.

Explanation of Solution

Given info: Proton is located at x = 6.00 fm.

From Conservation of energy, the kinetic energy is given by,

$KE=-\Delta PE$

Formula to calculate the kinetic energy is,

$KE=\frac{1}{2}{m}_{\alpha }{v}^2$

• $m_{\alpha }$ is the mass of alpha particle.
• v is the speed of alpha particle.

From the above equations,

$v=\sqrt{-\frac{\Delta PE}{2m_{\alpha }}}$

Substitute $-2.17\times {10}^{-13}\text{J}$ for $\Delta PE$ and $6.64\times {10}^{-27}\text{kg}$ for $m_{\alpha }$

$\begin{array}{c}v=\sqrt{-\frac{\left(-2.17\times {10}^{-13}\text{J}\right)}{6.64\times {10}^{-27}\text{kg}}}\\ =8.08\times {10}^6\text{m/s}\end{array}$

Conclusion:

The speed of alpha particle is $8.08\times {10}^6\text{m/s}$.

To determine

The speed of protons at infinity.

Answer to Problem 19P

The speed of proton at infinity is $1.24\times {10}^7\text{m/s}$.

Explanation of Solution

Given info: Proton is located at x = 6.00 fm.

Explanation:

The kinetic energy is equally split among the two protons.

$KE=\frac{P{E}_2}{2}$

Formula to calculate the kinetic energy is,

$KE=\frac{1}{2}{m}_p{v}^2$

• $m_p$ is the mass of proton.
• v is the speed of proton.

From the above equations,

$v=\sqrt{\frac{PE}{m_p}}$

Substitute $2.55\times {10}^{-13}\text{J}$ for $P{E}_2$ and $1.67\times {10}^{-27}\text{kg}$ for $m_p$

$\begin{array}{c}v=\sqrt{\frac{2.55\times {10}^{-13}\text{J}}{1.67\times {10}^{-27}\text{kg}}}\\ =1.24\times {10}^7\text{m/s}\end{array}$

Conclusion:

The speed of proton at infinity is $1.24\times {10}^7\text{m/s}$.