Answer :
Answer:
8.533 %
Explanation:
Energy of the electron is given by [tex]E= eV[/tex] 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 [tex]eV=\frac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\frac{2eV}{m}}[/tex]
Here V is the potential difference which is given as 1878.7197 V
m is the mass of electron [tex]=9.1\times 10^{-31}\ kg[/tex]
Charge of electron [tex]=1.6\times 10^{-19}[/tex]
Putting all these value in expression of velocity [tex]v=\sqrt{\frac{2eV}{m}}=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 1878.7197}{9.1\times 10^{-31}}}=25.7\times 10^6m/sec[/tex]
Speed of the light c =[tex]3\times 10^8m/sec[/tex]
So required percentage [tex]=\frac{25.6\times 10^6}{3\times 10^8}\times 100=8.533[/tex]
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,
[tex]\dfrac{1}{2}mv^2=eV[/tex]
[tex]v=\sqrt{\dfrac{2eV}{m}}[/tex]
e is the charge on electron
m is the mass of electron
[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1878.7197}{9.1\times 10^{-31}}}[/tex]
[tex]v=2.57\times 10^7\ m/s[/tex]
Speed of light, [tex]c=3\times 10^8\ m/s[/tex]
Required percentage of the speed of light,
[tex]\dfrac{2.57\times 10^7}{3\times 10^8}\times 100=8.56%\[/tex]
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.