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fomenos
4 years ago
7

An electron in an electron gun is accelerated from rest by a potential of 25 kV applied over a distance of 1 cm.

Physics
1 answer:
podryga [215]4 years ago
3 0

Answer:

9.38\times 10^7 m/s

Explanation:

We are given that

Potential ,V=25 kV=25\times 10^3 V

Distance,r =1 cm=\frac{1}{100}=0.01 m

1 m=100 cm

Mass of electron, m=9.1\times 10^{-31} kg

Charge, q=1.6\times 10^{-19} C

We have to find the final velocity of the electron.

Speed of electron,v=\sqrt{\frac{2qV}{m}}

Using the formula

v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 25\times 10^3}{9.1\times 10^{-31}}

v=9.38\times 10^7 m/s

Hence, the final velocity of the electron=9.38\times 10^7 m/s

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