The energy that should be added to an electron to reduce its de Broglie wavelength from one nm to 0.5 nm is

Explanation

Wavelength charge `= 2.25 - 1.75 = 0.5 Å`
Energy decreases `=(hc)/(lambda) = (6.626 xx 10^(-34) Js xx 3 xx 10^(6)ms^(-1))/(0.5 xx 10^(-10)m)`
`= 3.975 xx 10^(-15) J`
`(1.6 xx 10^(-19) C)v_(0) = 3.975 xx 10^(-15)`
`v_(0) = (3.975 xx 10^(-15))/(1.6 xx 10^(-19)) = 24843.75 V`
It looks very much ligical to do in this way .But the mistake here is that `lambda = c//v` is applicable to only electromagnetic wave and mol particle waves .The correct is as follow
`Delta E = (1)/(2) m (v_(1)^(2) - v_(2)^(2))`
`= (1)/(2) [((h)/( lambda_(1)))^(2) - ((h)/( m lambda _(2)))^(2)]`
`= (1)/(2) m(h^(2))/(m^(2)) [((1)/(lambda_(1)))^(2) - ((1)/(lambda_(2)))^(2)] = (h^(2))/(2m)[0.1289 xx 10^(19)]`
`eV_(0) = Delta E`
`V_(0) = (Delta E)/(e) = ((6.626 xx 10^(-34) J s)^(2) xx (0.1289 xx 10^(19) m^(-2)))/((2 xx 9.1 xx 10^(-31) kg)(1.6 xx 10^(-19) C))`
`= 1.943 V`