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

User Fayna
by
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2 Answers

5 votes

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.

User Lance Leonard
by
9.0k points
4 votes

Answer:

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

User WhiteTiger
by
9.1k points

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