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? , , 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 and proton system is given by:
$U_i = \frac{kq_eq}{r_i} = \frac{(9	imes10^9)(-1.6	imes10^{-19})(1.6	imes10^{-19})}{0.09} = -2.304	imes10^{-18}	ext{ J}$
where $q_e$ and $q_p$ are the charges of the electron and proton, respectively, $r_i$ is the initial separation between them, and $k$ is Coulomb's constant. Since the electron starts from rest, its initial kinetic energy is zero. Therefore, the total initial energy is $E_i = U_i$.

When the electron is 3.00 cm from the proton, the potential energy is given by:
$U_f = \frac{kq_eq}{r_f} = \frac{(9	imes10^9)(-1.6	imes10^{-19})(1.6	imes10^{-19})}{0.03} = -9.216	imes10^{-18}	ext{ J}$

At this point, the electron has both kinetic and potential energy, and conservation of energy requires that the total energy is conserved:
$E_f = \frac{1}{2}mv^2 - \frac{kq_eq}{r_f} = E_i$

Rearranging for $v$, we get:
$v = \sqrt{\frac{2(E_i + kq_eq/r_f)}{m}}$

Plugging in the values, we get:
$v = \sqrt{\frac{2((-2.304	imes10^{-18}) + (-9.216	imes10^{-18}))}{9.11	imes10^{-31}}} = 106	ext{ m/s}$

Therefore, the answer is B).

An Electron Is Released from Rest at a Distance of 9.00