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An electron (mass m(e ))falls through a ...

An electron (mass `m_(e )`)falls through a distance d in a uniform electric field of magnitude E.
,
The direction of the field is reversed keeping its magnitudes unchanged, and a proton(mass `m_(p)`) falls through the same distance. If the times taken by the electrons and the protons to fall the distance d is `t_("electron")` and `t_("proton")` respectively, then the ratio `t_("electron")//t_("proton")`.

A

1

B

`(m_p//m_e)^(1//2)`

C

`(m_e//m_p)^(1//2)`

D

1836

Text Solution

Verified by Experts

Let the distance to be traveled be `x`. Let the strength of uniform electric field be `E`. For the electron.
`u=0,S=x,a=(eF)/(m_(e)),t=t_(1)`
`S=ut+(1)/(2)at^(2)` or `x=(1)/(2)(eF)/(m_(e))xxt_(1)^(2)` (i)
For the proton,
`u=-0,s=x,a=(eF)/(mp),t=t_(1)`
`S=ut+(1)/(2)at^(2)` or `x=(1)/(2)(eF)/(m_(p))xxt_(2)^(2)` (ii)
From Eqs. (i) and (ii),
`(t_(2)^(2))/(t_(1)^(2))=(m_(p))/(m_(e))` or `(t_(2))/(t_(1))=[(m_(p))/(m_(e))]^(1//2)`
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