Question

For relativistic particles the de Broglie relation lambda_dB = h/p still holds, but recall that the relationship between the particle's momentum p and total energy E changes lambda_db = hc/squareroot K(K+2mc^2)

Answer 1

K = m c^2 - m0 c^2

and E = mc^2 = sqrt [ p^2 c^2 + m0^2 c^4]

and mc^2 = k + m0c^2

putting in,

k + m0c^2 = sqrt [ p^2 c^2 + m0^2 c^4]

square both sides,

k^2 + m0^2 c^4 + 2km0c^2 = p^2 c^2 + m0^2 c^4

k^2 + 2k m0c^2 = p^2 c^2

p = sqrt[ k (k + 2m0 c^2) ] / c

lambda = h / p

lambda = h / (sqrt[ k (k + 2m0 c^2) ] / c )

lambda = hc / sqrt[ k (k + 2m0 c^2) ]