A particle has a de Broglie wavelength of 3.30 x 10^-10 m. Then its kinetic energy quadruples. What is the particle's new de Broglie wavelength, assuming that relativistic effects can be ignored?

Answer 1

L = h/p = h/mv

v = h/mL
v^2 = h^2/(m^2 L^2)
m v^2 = h^2/(m L^2) (forget the 1/2, cancels)

new energy
m Vnew^2 = h^2/(m Lnew^2) = 4 h^2/(m L^2)
so
1/Lnew^2 = 4/L^2
4 Lnew^2 = L^2
2 Lnew = L
Lnew = (1/2) L

1.65*10^-10 m