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If an electron moving with a speed of `2.5xx10^(7)ms^(-1)` is deflected by an electric field of `1.6k(V)m^(-1)` perpendicular to its circular path, then `(e)/(m)` for the electron will be (given radius of circular path =2.3m):

A

`1.67xx10^(11)Ckg^(-1)`

B

`1.76xx10^(11)Ckg^(-1)`

C

`1.7xx10^(11)Ckg^(-1)`

D

`1.59xx10^(11)Ckg^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C

`eE=(mv^(2))/(r)`
`:.(e)/(m)=(v^(2))/(Er)=1*7xx10^(11)C Kg^(-1)`
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