A helium ion of mass 4m and charge 2e is accelerated from rest through a potential difference V in vacuum. Its final speed will be

Question

A helium ion of mass 4m and charge 2e is accelerated from rest through a potential difference V in vacuum. Its final speed will be

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Linh Đan 5 years 2021-09-01T16:37:37+00:00 1 Answers 295 views 1

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  1. Nick
    0
    2021-09-01T16:39:30+00:00
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    Answer:

    v=\sqrt{\frac{ev}{m} }

    Explanation:

    The helium ion has:

    a potential difference in vacuum = V,

    Charge = 2e

    and mass = 4m,

    speed = v,

    mass = 4m

    From electrostatics, the work done is the product of charge and its potential difference.

    Therefore, Work done = charge × potential difference = 2e × V = 2eV

    This work done is in form of kinetic energy, therefore:

    Kinetic energy = 1/2 × mass × speed²

    ⇒  Work done = Kinetic energy

    2eV=\frac{1}{2} *4m*v^2=2m*v^2\\v^2=\frac{2eV}{2m}\\ v^2=\frac{eV}{m}\\ v=\sqrt{\frac{ev}{m} }

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