- #1
JD_PM
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I was reading *Introduction to Nuclear Physics* by Krane and stumbled on the following (page 47):
In Elastic scattering, the initial electron wave function is of the form ##e^{i k_i r}## (free particle of momentum ##p_i = \hbar k_i##). The scattered electron can also be regarded as a free particle of momentum ##p_f = \hbar k_f## and wave function ##e^{i k_f r}##.
The interaction ##V(r)## converts the initial wave into the scattered wave; the probability for the transition will be proportional to the square of the following quantity:
$$F(q) = \int V(r) e^{iqr}dv$$
Plugging both Coulomb's potential and charge-per-unit-volume into ##F(q)##:
$$F(q) = \int e^{iqr'} \rho(r') dv'$$
Normalizing and knowing that ##\rho(r')## just depends on ##r'## (and not on ##\theta'## nor ##\phi'##) we get:
$$F(q) = \frac{4\pi}{q}\int r' sin (qr') \rho(r') dr'$$
Where ##q = k_i - k_f##. The scattering is elastic, so momentum is conserved (##p_i = p_f##) and ##q## is merely a function of the scattering angle ##\alpha## between ##p_i## and ##p_f##.
**Now a bit of vector manipulation shows:**
$$q = \frac{2p}{\hbar}sin(\frac{\alpha}{2})$$
**I do not know how to get the last expression**
In Elastic scattering, the initial electron wave function is of the form ##e^{i k_i r}## (free particle of momentum ##p_i = \hbar k_i##). The scattered electron can also be regarded as a free particle of momentum ##p_f = \hbar k_f## and wave function ##e^{i k_f r}##.
The interaction ##V(r)## converts the initial wave into the scattered wave; the probability for the transition will be proportional to the square of the following quantity:
$$F(q) = \int V(r) e^{iqr}dv$$
Plugging both Coulomb's potential and charge-per-unit-volume into ##F(q)##:
$$F(q) = \int e^{iqr'} \rho(r') dv'$$
Normalizing and knowing that ##\rho(r')## just depends on ##r'## (and not on ##\theta'## nor ##\phi'##) we get:
$$F(q) = \frac{4\pi}{q}\int r' sin (qr') \rho(r') dr'$$
Where ##q = k_i - k_f##. The scattering is elastic, so momentum is conserved (##p_i = p_f##) and ##q## is merely a function of the scattering angle ##\alpha## between ##p_i## and ##p_f##.
**Now a bit of vector manipulation shows:**
$$q = \frac{2p}{\hbar}sin(\frac{\alpha}{2})$$
**I do not know how to get the last expression**