The potential barrier reduces to `4.05 xx 10^(-2)` V in a diode and an electron with a speed `v_(1) = 1.3 xx 10^(5) ms^(-1)` is made to enter p-side from n-side. The velocity of the electron in p-side will be given as ______ `xx 10^(4) ms^(-1)`. (Take mass of electron `= 9 xx 10^(-31)` kg and charge of electron = `1.6 xx 10^(-19)` C)

Answer

`05.00`

Explanation

Given: `V = 4.05 xx 10^(-2) V, v_(1) = 1.3 xx 10^(3) m//s`
According to law of conservation of energy,
`1/2 mv_(1)^(2) - 1/2 mv_(2)^(2) = eV`
`therefore v_(2)^(2) = v_(1)^(2) - (2eV)/m`
`therefore v_(2)^(2) = (1.3 xx 10^(5))^(2) - (2 xx 1.6 xx 10^(-19) xx 4.05 xx 10^(-2))/(9 xx 10^(-31))`
`= 1.61 xx 10^(10) - (2 xx 8 xx 0.45 xx 10^(10))/5`
`= 0.25 xx 10^(10)`
`therefore |V_(2)| = 0.5 xx 10^(5) = 5 xx 10^(4) m//s`