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Suppose the an electron in the picture t...

Suppose the an electron in the picture tube of a televisoin set is accelerated from rest through a potential differnece `V_(b)-V_(a)= V_(ba) = +5000V`.
(a) What is the change in electric potential energy of the electron?
(b) What is the speed of the electron `( m = 9.1 xx 10^(-31) kg)` as a result of this acceleration ?

Text Solution

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(a) `DeltaU = -5000 eV `
`=-5000xx1.6xx10^(-10)J`
`=-8.0xx10^(-16) J.`
(b) `Delta K = K -0 = (1)/(2) mv^(2)= 8xx10^(-16)` J
`implies v=sqrt((2xx8xx10^(-16))/(1.67xx10^(-27)))=sqrt((1.60)/(1.67)xx10^(10))=9.8 xx10^(5)` m/s
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