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An electron is accelerated in an electri...

An electron is accelerated in an electric field of `40V cm^(-1)`. If `e//m` of electron is `1.76 xx10^(11) Ckg^(-1)` , then its acceleration is

A

`8.8 xx10^(14) m//sec^2`

B

`6.2 xx10^(13) m//sec^2`

C

`5.4 xx10^(12) m//sec^2`

D

Zero

Text Solution

Verified by Experts

The correct Answer is:
A
`ma = qEimplies a =(qE)/(m) = 1.76 xx10^(11) xx50 xx10^2`
`=8.8 xx10^(14)m//sec^2`
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Knowledge Check

  • An electron is accelerated through a potential difference of 200 volts. If e//m for the electron be 1.6 xx 10^(11) coulomb/kg, the velocity acquired by the electron will be

    A
    `8 xx 10^(6) m//s`
    B
    `8 xx 10^(5)m//s`
    C
    `5.9 xx 10^(6)m//s`
    D
    `5.9 xx 10^(5)m//s`

  • Maximum velocity of photoelectron emitted is 4.8 ms^(-1) . If e/m ratio of electron is 1.76xx10^(11)Ckg^(-1) , then stopping potential is given by

    A
    `5xx10^(-10)` J/C
    B
    `3xx10^(-7)` J/C
    C
    `7xx10^(-11)` J/C
    D
    `2.5xx10^(2)` J/C

  • If the electronic charge e=1.6xx10^(-19)C and the specific charge e/m off electron is 1.76xx10^(11)C(kg)^(-1) then calculate the mass m of the electron.

    A
    `3.2xx10^(-31)kg`
    B
    `5.1xx10^(-31)kg`
    C
    `9.1xx10^(-31)kg`
    D
    `8.2xx10^(-31)kg`