Question

Question #5669f

Answer 1

`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` ①
- Electron Wavelength (de Broglie equation): `λ = h/(mv)` ②
- Electron Kinetic Energy: `E_k = 1/2 mv^2` ③

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 ))` ④

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)` ⑤

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`