Question

a proton gun. you and your team are designing a device that creates a beam of high-velocity protons. the device consists of a proton source that emits protons initially at rest at the positive plate. the protons accelerate toward the negative plate, which has a hole that allows the proton to pass through it. the plates are separated by 6.0 cm, and the potential difference between the plates is 4.5 kv.(a) what is the kinetic energy of the protons when they first through the hole?(b) what is the speed of the protons when they pass through the hole?(c) what is the acceleration of the proton between the plates?(a) number □ units □(b) number □ units □(c) number □ units □

Explanation:

Step1: Relate KE to electric potential energy

The kinetic energy gained by a proton equals the electric potential energy lost, so $KE = qV$. The charge of a proton $q = 1.6\times10^{-19}\ \text{C}$, potential difference $V = 4.7\ \text{kV} = 4700\ \text{V}$.

Step2: Calculate kinetic energy

Substitute values into the formula:
$KE = 1.6\times10^{-19}\ \text{C} \times 4700\ \text{V} = 7.52\times10^{-16}\ \text{J}$

Step3: Relate KE to speed

Use the kinetic energy formula $KE = \frac{1}{2}mv^2$, rearrange to solve for $v$: $v = \sqrt{\frac{2KE}{m}}$. Mass of proton $m = 1.67\times10^{-27}\ \text{kg}$.

Step4: Calculate the speed

Substitute values into the speed formula:
$v = \sqrt{\frac{2\times7.52\times10^{-16}\ \text{J}}{1.67\times10^{-27}\ \text{kg}}} \approx \sqrt{9.018\times10^{11}} \approx 9.496\times10^{5}\ \text{m/s}$

Step5: Relate electric field to potential difference

Use $E = \frac{V}{d}$, where $d = 0.06\ \text{m}$.

Step6: Calculate electric field

Substitute values into the field formula:
$E = \frac{4700\ \text{V}}{0.06\ \text{m}} \approx 78333.33\ \text{V/m}$

Answer:

a) $7.52\times10^{-16}\ \text{J}$
b) $9.50\times10^{5}\ \text{m/s}$ (rounded to 3 significant figures)
c) $7.83\times10^{4}\ \text{V/m}$ (rounded to 3 significant figures)