Question

What amount of energy (in eV) should be added to an electron to reduce its de-Broglie its de-Broglie wavelength from 100 to 50 pm? Given `h=6.62xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg`.

Answer 1

From `lambda=(2pi ħ)/(P)=(2pi ħ)/(sqrt(2mT))`
we find `T=(4pi^(2) ħ^(2))/(2 m lambda^(2))=(2pi^(2) ħ^(2))/(m lambda^(2))` Thus `T_(2)-T_(1)=(2pi^(2) ħ^(2))/(m)((1)/(lambda_(2)^(2))-(1)/(lambda_(1)^(2)))`
Substitution gives `Delta T= 451 eV= 0.451ke V`.