Question

Find the energy that should be added to an electron of energy 2eV to reduce its de-Broglie wavelength from 1nm to 0.5nm.

Answer 1

Correct Answer - `6 eV`
As `lambda=h/(sqrt(2mK)) so K=(h^(2))/(2mlambda^(2)) or K prop 1/(lambda^(2))`
`:. (K_(1))/(K_(2))=[(lambda_(2))/(lambda_(1))]^(2)=[0.5/1]^(2)=1/4`
or `K_(2)=4K_(1)=4xx2eV=8eV`
Increases in energy `=K_(2)-K_(1)=8-2=6eV`