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If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed?

e=1/60 x 10^-19 C
K=8.99 x10^9
Me=9.11 x 10^-31

Answer :

Answer:

Speed of the electron will be [tex]v=5.896\times 10^7m/sec[/tex]

Explanation:

We have given that charge on electron [tex]e=1.6\times 10^{-19}C[/tex]

Mass of electron [tex]m=9.11\times 10^{-31}kg[/tex]

Potential difference = [tex]V=9.9KV=9.9\times 10^3volt[/tex]

Now according to energy conservation [tex]eV=\frac{1}{2}mv^2[/tex]

[tex]1.6\times 10^{-19}\times 9.9\times 10^3=\frac{1}{2}\times 9.11\times 10^{-31}v^2[/tex]

[tex]v=5.896\times 10^7m/sec[/tex]

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