Find the speed of an elctron with kinetic energy (a) 1 eV , (b) 10 KeV and (c ) 10 MeV.

Explanation

(a) m_0c^(2) / 2 (sqrt - 1 v^(2) c^(2) ) - m_0c^(2) = 1.6 xx 10^(-19)`
`rArr m_0c^(2) (1 / (sqrt 1 -v^(2) / c^(2)) - 1)) = 1.6 xx 10^(-19) / 9.1 xx 10^(-31) xx 9 xx 10^(16) ` rArr 1 / (sqrt 1- v / c^(2)) -1 =
` 1.6 xx 10^(-19) / 9.1 xx 10^(-31) xx 9xx 10^(16)`
` rArr 1 / (sqrt - v/ c^(2)) - 1 = 0.019536 xx 10^(-4)`
` rArr 1 / (sqrt 1 -v/ c^(2)) = 1 + 0.019536 xx 10^(-4) `
` = 1.0000019536 `
`rArr 1 / (sqrt 1 -v/ c^(2)) = 1 / 1.0000019536 `
` rArr 1 -v^(2) / c^(2) = 0.99999613 `
` v^(2) / c^(2) = 0.00000387 `
` v/ c = 0.001967231 `
` v = c xx 0.001967231 `
` = 3 xx 0.001967231 xx 10^(8) `
` = 5.92 xx 10^(5)m/ s `
(b) ` m_0 c^(2) / 2 (sqrt 1 -v^(2)/c^(2) ) -1) - m_0c^(2)`
` `= 1.6 xx 10^(-19) xx 10 xx 10^(3)`
m_0c^(2) (1 / (sqrt 1 -v^(2) / c^(2) ) -1) = 1.6 xx 10^(-15)`
` rArr 9.1 xx 10^(-31) xx 9 xx 10^(16)(1 / (sqrt 1 -v^(2) /c^(2))- 1) `
` = 1.6 xx 10^(-15) `
`rArr 1 / (sqrt 1 -v^(2) / c^(2) ) - 1 = 1.6 xx 10^(15) / 9.1 xx 9 xx 10^(-15) `
` rArr 1 / (sqrt 1-v^(2)/c^(2)) - 1 = 1.6 / xx 9.1 xx 9`
` rArr 1 / (sqrtv^(2) /c^(2) - 1 = 1.6 xx 10^(-15) / 9.1 xx 9 xx 10^(-15) `
` rArr 1 / (sqrt 1-v^(2) / c^(2) ) - 1 = 1.6 / 9.1 xx 9`
` rArr 1 / (sqrt 1 - v^(2)/ c^(2)) = 0.980838 `
` 1 -v^(2) / c^(2) = 1- 0.962043182 `
` = 0.037956817 `
` v^(2) = 0.341611359 xx 10^(18) `
` v = 0.584475285 xx10^(8) `
` = 10^(7)m/s`
`K.E. = 10MeV`
` = 109 xx 10^(6)eV `
` = 10^(7) xx 1.6 xx 10^(-19)J`
` = 1.6 xx 10^(-12) J`
`rArr m_0c^(2) / 2(sqrt 1 -v^(2)/c^(2)) - m_0c^(2) = 1.6 xx 10^(-12)J `
` rArr m_0c^(2) (1) / (sqrt 1 - v^(2) / c^(2) ) - 1)) = 1.6 xx 10^(-12) J`
`rArr 1 / 1-v^(2) / c^(2)) - 1 = 1.6 xx 10^(-12) / 9.1 xx 9 xx 10^(-31) xx 10^(16) `
` rArr 1 / (sqrt 1 -v^(-2) / c^(2)) - 1 = 1.019536 xx 10^(3) `
` rArr 1 / (sqrt 1 -v^(2)c^(2)) - 1 = 1019.536 + 1`
` rArr (sqrt 1-v^(2) / c^(2)) = 0.000979877 `
` rArr 1-v^(2)/c^(2) = 0.99 xx 10^(-6)`
` v = 2.999999039 xx 10^(8)m/sec.`