Where does the $0.61$ come from in $ R \sin \theta = 0.61\lambda $? In my A2 Physics textbook it says that 
In the estimation of the nuclear radius by using electron diffraction the equation 
$$ R \sin \theta = 0.61\lambda  $$
where:
$R$ = radius of the nucleus
$\theta$ = the angle of diffraction 
What I cannot understand and what my textbook does not mention is where the 0.61 part comes from. How is this derived and what is the significance of $0.61$?
 A: 
What I cannot understand and what my textbook does not mention is where the 0.61 part comes from. How is this derived and what is the significance of 0.61?

In your equation if  you take nuclear diameter $D=2R$
Then your relation  looks like $ D\sin\theta = 1.22 \lambda$
If one wishes to resolve two objects separated by a distance $D$, one can use Rayleigh Criterion. (Luboš Motl comments can also help you)
By Rayleigh Criterion of limit of resolution  (as an approximate condition) we say  that two objects  are too close together to resolve (identify as two separate particles rather than one) if their central diffraction peaks overlap    such that the first order minimum of one falls  on the central maximum of the other one.
This leads to dip in the maximum difracted intensity and first appearance of two objects takes place (this we usually employ in spectral resolution of two lines of sodium doublet).
Since the angle spanned by the central diffraction peak is $1.22\lambda/D$, this means that we say we can't resolve objects unless the angular separation is greater than $1.22\lambda/D$.
1.22 sounds like a pretty strange number. It turns out that the intensity of diffracted light from a circular aperture  is given by
$$I(\theta) = I_0 \left( 2 \frac{J_1(x)}{x} \right)^2$$
here  $J_1(x)$ is the "Bessel function of the first kind, of order one",  $x=kD \sin \theta$    and  $k=2\pi/\lambda$. 
$J_1(x)$  is a special function that turns up in  in the solutions to partial differential equations  depicting the scattering,  especially in partial wave analysis.
 One  would like to know when the Bessel function goes to zero. since the first place where the intensity is zero is the edge of the central diffraction spot.
the first zero of Bssel function of order 1 is at approx. x=3.84 

see  http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html

