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

Question

An electron in an electron gun is accelerated from rest by a potential of 25 kV applied over a distance of 1 cm.
The final velocity of the electrons is _____.
The mass of the electron is 9.1×10^(-31) kg and its charge is 1.6×10^(-19) C.

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MichaelMet 2 months 2021-07-31T16:48:38+00:00 1 Answers 5 views 0

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    2021-07-31T16:49:38+00:00

    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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