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Take the particle in question number 49 ...

Take the particle in question number 49 an electron projected with velocity `v_(x)=4xx10^(6)ms^(-1)`. If electric field between the plates separated by 1 cm is `8.2xx10^(2)NC^(-1)`, then the electron will strike the upper plate if the length of plate is (Take `m_(e)=9.1xx10^(-31)kg`)

A

2.14cm

B

3.9cm

C

1.23cm

D

3.3cm

Text Solution

Verified by Experts

The correct Answer is:
D
Given `v_(x)=4xx10^(6)ms^(-1),d=1cm=1xx10^(-2)m`
`E=8.2xx10^(2)NC^(-1),q=e=1.6xx10^(-19)C,m_(e)=9.1xx10^(-31)kg`
The electron will strike the upper plate at its other end of x = L as soon as its deflection.
And `y=d/2=10^(-2)/2m=5xx10^(-3)m`
From eqn. (iii),
`L=sqrt((2m_(e)v_(x)^(2)y)/(qE))=sqrt((2xx9.1xx10^(-31)xx(4xx10^(6))^(2)xx5xx10^(-3))/(1.6xx10^(-19)xx8.2xx10^(2)))`
`L=3.3xx10^(-2)m` = 3.3 cm
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