The energy that should be added to an electron to reduce its de - broglie wavelength from 1 nm to 0.5 nm is

Options

A. Four -0 times the initial energy
B. equal to the initial energy
C. Two - times the initial energy
D. Three - times the initial energy

Answer

D

Explanation

We know that
We have ` lamda = (h) /( sqrt(2 mk))`
` therefore ` kinetic energy (KE) ` prop (1)/( lamda ^2)`
hence
` (KE_2)/(KE_1) =((lamda_1)/(lamda_2))^2`
here `lamda _1 = 1nm`
` lamda _2 ` =0.5
` (KE _2)/(KE_1)=((1)/(0.5 ))^2`
` KE _2 = 4 KE _1`
hence , the kinetic energy increases three times .
` Delta KE = 3 KE_1`