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An alpha particle and a proton fall thro...

An `alpha` particle and a proton fall through the same potential difference. Find the ratio of their momentra if the mass of the `alpha` particle is four times that of the proton.

Text Solution

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(i) Compare the velocity of an `alpha` particle with that of a proton in the same potential difference. To compare the velocities equate the kinetic energy. We know that momentum p = mv.
Compare mass and velocity of `alpha` particle with that of proton to find the relation between their momebta.
(ii) `sqrt(8) : 1`
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Knowledge Check

  • A proton and an alpha - particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and alpha -particle is

    A
    `sqrt2`
    B
    `(1)/(sqrt2)`
    C
    `2sqrt2`
    D
    None of these

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    `sqrt(8)`
    D
    `(1)/(sqrt(8))`

  • The ratio of the e/m values of a proton and an alpha -particle is:

    A
    `2:1`
    B
    `1:1`
    C
    `1:2`
    D
    `1:4`

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