An electron is released from rest at a distance of 9.00 cm from a proton.If the proton is held in place,how fast will the electron be moving when it is 3.00 cm from the proton? (m_el= 9.11 × 10^-31 kg,e = 1.60 × 10^-19 C,k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²) A)75.0 m/s B)106 m/s C)130 m/s D)1.06 × 10³ m/s E)4.64 × 10⁵ m/s

Question

Quizplus · Accepted Answer

We can use conservation of energy to solve this problem. The initial potential energy of the electron-proton system is given by:
\begin{align*}
U_i &= -\frac{kq^2}{r_i} \
&= -\frac{(8.99	imes10^9	ext{ N}\cdot	ext{m}^2/	ext{C}^2)(1.60	imes10^{-19}	ext{ C})^2}{0.09	ext{ m}} \
&= -2.307	imes10^{-18}	ext{ J} \
\end{align*}
At a distance of 3.00 cm from the proton, the potential energy is:
\begin{align*}
U_f &= -\frac{kq^2}{r_f} \
&= -\frac{(8.99	imes10^9	ext{ N}\cdot	ext{m}^2/	ext{C}^2)(1.60	imes10^{-19}	ext{ C})^2}{0.03	ext{ m}} \
&= -6.545	imes10^{-18}	ext{ J} \
\end{align*}
Thus, the initial potential energy is converted into kinetic energy as the electron moves towards the proton:
\begin{align*}
\Delta K &= -\Delta U \
&= U_f - U_i \
&= 4.238	imes10^{-18}	ext{ J} \
\end{align*}
Since the electron is initially at rest, all of the kinetic energy is converted into kinetic energy at a distance of 3.00 cm:
\begin{align*}
K_f &= \Delta K \
&= 4.238	imes10^{-18}	ext{ J} \
\end{align*}
The kinetic energy can also be expressed in terms of the speed of the electron:
\begin{align*}
K &= \frac{1}{2}mv^2 \
v &= \sqrt{\frac{2K}{m}} \
\end{align*}
Substituting the values for K and m, we get:
\begin{align*}
v &= \sqrt{\frac{2(4.238	imes10^{-18}	ext{ J})}{9.11	imes10^{-31}	ext{ kg}}} \
&= 106	ext{ m/s} \
\end{align*}
Therefore, the electron will be moving at 106 m/s when it is 3.00 cm from the proton, which is answer choice B.

An Electron Is Released from Rest at a Distance of 9.00