# Question 5669f

Aug 31, 2017

$E = \frac{2 c m l a m \mathrm{da}}{h} {E}_{k}$

#### 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: $E = \frac{h c}{\lambda}$
- Electron Wavelength (de Broglie equation): λ = h/(mv)
- Electron Kinetic Energy: ${E}_{k} = \frac{1}{2} m {v}^{2}$

Rearrange equation 2 to get an expression for $m {v}^{2}$ 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
${E}_{k} = \frac{1}{2} \left({h}^{2} / \left(m {\lambda}^{2}\right)\right)$

Rearrange equation 1 to get an expression for the wavelength
⇒ lambda =(hc)/E
Substitute that into equation 4
${E}_{k} = \frac{1}{2} \left({h}^{2} / \left(m {\left(\frac{h c}{E}\right)}^{2}\right)\right) = \frac{1}{2} {E}^{2} / \left(m {c}^{2}\right)$

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.
$v = \frac{h}{m l a m \mathrm{da}}$
⇒ E = (2c)/(h/(mlambda)) E_k 

⇒ E = (2cmlamda)/h E_k#