Question #5669f

1 Answer
Aug 31, 2017

Answer:

# E = (2cmlamda)/h E_k#

Explanation:

I didn't get the answer of #(2 π mc)/h# times the KE, instead I got #(2 λ mc)/h#, but I hope the method is helpful. It is possible that there are one or two shortcuts that I missed.

My starting point is the following three equations from quantum physics:
- Photon Energy: #E=(hc)/lambda#
- Electron Wavelength (de Broglie equation): #λ = h/(mv)#
- Electron Kinetic Energy: #E_k = 1/2 mv^2#

Rearrange equation 2 to get an expression for #mv^2# so that we can substitute that into equation 3.
#mv = h/lambda ⇒ mv^2 = (mv)^2 /m = h^2 /(mlambda^2 )#

Now substitute that into equation 3
#E_k = 1/2 (h^2 /(mlambda^2 ))#

Rearrange equation 1 to get an expression for the wavelength
#⇒ lambda =(hc)/E#
Substitute that into equation 4
#E_k = 1/2 (h^2 /(m((hc)/E)^2 )) = 1/2 E^2 / (mc^2)#

Make wavelength from equations 1 and 2 equal to each other to eliminate wavelength from the equations:
#λ = (hc)/E = h/(mv) ⇒ E = mcv#

Replace #E²# with #E × E# in equation 5 and substitute for one of the E's with the above expression.
#E_k = 1/2 (E × mcv) / (mc²) ⇒ E = (2c)/v E_k#

Lastly use equation 2 to get an expression for v and substitute that into the above equation.
# v = h/(mlamda)#
#⇒ E = (2c)/(h/(mlambda)) E_k #

#⇒ E = (2cmlamda)/h E_k#