Question

De Broglie postulated that the relationship ? = h/p is valid for relativistic particles. What is the de Broglie wavelength for a (relativistic) electron having a kinetic energy of 3.39 MeV?

answer in m

Answer 1

(1/2)mv^2 = qV =KE

v = sqrt(2qV/m) = 1.03 x 10^8 m/s

wavelength = h /p= 6.6 x 10^-34/2.48*10^-12=2.6556*10^-22 M

Answer 2

There is a relation- P(momentum)=sqrt(2mK) (root of 2 X mass X Kinetic enegy). convert 3.25 MeV to joules first and apply in Eqn

P=9.73*10^-22

lambda= 680.9 nm

Answer 3

There is a relation- P(momentum)=sqrt(2mK) (root of 2 X mass X Kinetic enegy). convert 3.25 MeV to joules first and apply in Eqn

P=9.73*10^-22

lambda= 6.809*10^-13 m

Answer 4

-Electron has 3.39 MeV (or 4.8*10^-13 Joules)
-it's relativistic
-finding ?.

h=6.63*10^-34

?=h/p (obviously)

And I'm not sure if they're needed, but the relativistic eq's are:

KE = mc^2/sqrt(1-(v/c)^2)
p = mv/sqrt(1-(v/c)^2)

E = hc/?

4.8E-13 = (6.63E-34)(3E8)/?
? = (6.63E-34)(3E8)/(4.8E-13)
? = 4.14E-13 m