Question

What percentage of the speed of light will an electron be moving after being accelerated through a potential difference of 1,878.7197 Volts?

Answer 1

Step-by-step explanation:

Potential difference, V = 1878.7197 volts

Energy gained by the electron when it is accelerating is, eV

Let v is the velocity and m is the mass of the electron. So,

`(1)/(2)mv^2=eV`

`v=\sqrt{(2eV)/(m)}`

e is the charge on electron

m is the mass of electron

`v=\sqrt{(2* 1.6* 10^(-19)* 1878.7197)/(9.1* 10^(-31))}`

`v=2.57* 10^7\ m/s`

Speed of light,
`c=3* 10^8\ m/s`

Required percentage of the speed of light,

`(2.57* 10^7)/(3* 10^8)* 100=8.56%\`

So, the electron be moving after being accelerated through a potential difference of 1,878.7197 volts is 8.56 % of the speed of light.

Answer 2

8.533 %

Step-by-step explanation:

Energy of the electron is given by
`E= eV` where e is charge of electron and V is the potential difference, this energy will accelerate the electron and so this energy will be converted into kinetic energy

So
`eV=(1)/(2)mv^2`

`v=\sqrt{(2eV)/(m)}`

Here V is the potential difference which is given as 1878.7197 V

m is the mass of electron
`=9.1* 10^(-31)\ kg`

Charge of electron
`=1.6* 10^(-19)`

Putting all these value in expression of velocity
`v=\sqrt{(2eV)/(m)}=\sqrt{(2* 1.6* 10^(-19)* 1878.7197)/(9.1* 10^(-31))}=25.7* 10^6m/sec`

Speed of the light c =
`3* 10^8m/sec`

So required percentage
`=(25.6* 10^6)/(3* 10^8)* 100=8.533`