A free particle with initial kinetic energy E, de-Broglie wavelength `lambda`, enters a region where in it has a potential V, what is the new de-Broglie wavelength?

Options

A. `lambda(1-V//E)`
B. `lambda(1+E//V)`
C. `lambda//(1-V//E)^(2)`
D. `lambda//(1+V//E)^(2)`

Answer

c

Explanation

When a particle of kinetic energy `E` goes to a region where it has now a potential energy `V`, then its new kinetic energy, `E'=E-V`, because the energy `V` has been taken out of its K.E. similar to a body going on a height losing K.E. and gaining P.E.
De-Broglie wavelength, `lambda=h/(sqrt(2mE))`
`lambda'=h/(sqrt(2mE'))=h/(sqrt(2m(E-V)))`
`=h/(sqrt(2mE(1-V//E)))=lambda/(sqrt((1-V//E)))`