Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 114 V. m/s (b) Calculate the speed of an electron that is accelerated through the same potential difference. m/s

A) v = 148,242.72 m/s

B) v = 6,328,025.58 m/s

Explanation:

To solve this, we will equate electric potential to kinetic energy.

Formula for Electric potential is qV where q is charge and V is potential difference.

While formula for kinetic energy is ½mv² where m is mass and v is velocity

Thus;

qV = ½mv²

Let us make the velocity the formula;

v = √(2qV/m)

A) PROTON

Charge of proton has a constant value of 1.6 × 10^(-19) C

Mass of proton has a constant value of 1.66 × 10^(-27) kg

We are given that potential difference = 114 V.

So, v = √(2qV/m)

Thus; v = √(2*1.6 × 10^(-19)*114/(1.66 × 10^(-27)))

v = 148,242.72 m/s

B) ELECTRON

Charge of electron has a constant value of 1.6 × 10^(-19) C

Mass of electron has a constant value of 9.11 × 10^(-31) kg

v = √(2qV/m)

Thus;

v = √(2*1.6*10^(-19)*114)/(9.11 × 10^(-31)))

v = 6,328,025.58 m/s

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge