# Determining Form factor from density distribution

• Rayan
Rayan
Homework Statement
Show that the form factor of density distribution $$\rho (r)$$ is $$F(q^2)$$
Relevant Equations
$$\rho (r) = \rho_0 \cdot e^{-\frac{r}{R}}$$
$$F(q^2) = \frac{8\pi \rho_0 R^3}{1 + \frac{q^2R62}{h^2} }$$
So my first thought was that I can just use Fourier trick and integrate:

$$F(q^2) = \int_V \rho(r) \cdot e^{ i \frac{ \vec{q} \cdot \vec{r} }{h} } d^3r$$

$$F(q^2) = 2\pi \rho_0 \int_0^{\infty} r^2 \cdot e^\frac{-r}{R} dr \cdot \int_0^{\pi} \sin{\theta} \cdot e^{ -i \frac{q \cdot r \cos(\theta) }{h} } d\theta$$

$$F(q^2) = 2\pi \rho_0 \frac{-h}{iq} ( \frac{ e^{ i \frac{q \cdot r }{h} } - e^{ -i \frac{q \cdot r }{h} } }{r} ) \int_0^{\infty} r^2 \cdot e^{\frac{-r}{R}} dr$$

$$F(q^2) = \frac{-4\pi h \rho_0}{q} \int_0^{\infty} \sin(\frac{qr}{h}) e^{\frac{-r}{R}} \cdot r dr$$

But the integral is very complicated, which probably means I missed up somewhere on the way, but I can't really see it! Any tips?

Last edited:
Rayan said:
$$F(q^2) = 2\pi \rho_0 \frac{-h}{iq} ( \frac{ e^{ i \frac{q \cdot r }{h} } - e^{ -i \frac{q \cdot r }{h} } }{r} ) \int_0^{\infty} r^2 \cdot e^{\frac{-r}{R}} dr$$
The expression in parentheses is a function of ##r## and should be inside the integral. Instead of combining the two exponentials into a sine function, you might try leaving them as exponential functions. Can you work out the following integral? $$\int_0^{\infty} e^{ i \frac{q \cdot r }{h}}\cdot e^{\frac{-r}{R}} rdr$$

Rayan
TSny said:
The expression in parentheses is a function of ##r## and should be inside the integral. Instead of combining the two exponentials into a sine function, you might try leaving them as exponential functions. Can you work out the following integral? $$\int_0^{\infty} e^{ i \frac{q \cdot r }{h}}\cdot e^{\frac{-r}{R}} rdr$$
You're totally right! I managed to solve this integral instead and got the right answer! Thank you so much!!:)

TSny

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