Chapter 17, Problem 55GP

(a)

To determine

The kinetic energy acquired by proton.

Answer to Problem 55GP

The proton gains the same kinetic energy, $6.3keV$

Explanation of Solution

Introduction:

(Kinetic energy) KE of an object is energy that it possesses due to its motion. In other words, the energy of motion is known s kinetic energy. Kinetic energy depends on mass and speed and also, it is measured Joules.

It is given that an electron starting from rest acquires $6.3keV$ of KE in moving from the point $A$ to $B$ . Means that electron was accelerated through a potential difference $-6.3kV$ in gaining $6.3kV$ of kinetic energy.

The proton is accelerated through the opposite potential difference (PD) as an electron. Also, it has exact opposite charge.

So, the proton will gain the same KE.

(b)

To determine

Ratio of the speed.

Answer to Problem 55GP

  $\text{42}\text{.8}$

Explanation of Solution

Given data:

Proton and electron have the same kinetic energy.

Mass of proton is $\text{1}{\text{.67×10}}^{\text{-27}}\text{kg}$

Mass of electron is $\text{9}{\text{.11×10}}^{\text{-31}}\text{kg}$

Formula used:

The energy of motion is known s kinetic energy. The formula of kinetic energy is $\text{KE = }\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}$

Where,

  $\text{m}$ is mass $\left(\text{kg}\right)$

  $\text{v}$ is velocity $\left(\text{m/s}\right)$

  $\text{KE}$ is kinetic energy $\left(\text{J}\right)$

Calculation:

According to the question, electron and proton has the same kinetic energy. To find the ratio of the speed, use the formula of kinetic energy.

  $\frac{\text{1}}{\text{2}}{\text{m}}_{\text{p}}{\text{v}}_{\text{p}}^{\text{2}}\text{=}\frac{\text{1}}{\text{2}}{\text{m}}_{\text{e}}{\text{v}}_{\text{e}}^{\text{2}}$

  $\frac{{\text{v}}_{\text{e}}}{{\text{v}}_{\text{p}}}\text{=}\sqrt{\frac{{\text{m}}_{\text{p}}}{{\text{m}}_{\text{e}}}}\text{=}\sqrt{\frac{\text{1}{\text{.67×10}}^{\text{-27}}\text{kg}}{\text{9}{\text{.11×10}}^{\text{-31}}\text{kg}}}\text{=42}\text{.8}$