# Electric Quadrupole Moment of the Deuteron. 1/4 factor

Sometimes in books I use to see this expression $$\widehat{Q}_{0}=\left(3 z^{2}-r^{2}\right)=\sqrt{\frac{16 \pi}{5}} r^{2} Y_{20}(\theta, \phi)$$ for quadrupole electric moment of the deuteron but sometimes i can see $$\widehat{Q}_{d}=\frac{1}{4} \widehat{Q}_{0}$$ I know that $$1/4$$ factor it could be related with the center of mass reference frame but I don't understand how to get it. The thing is that in some books explain it from $$r_{d}=\frac{1}{4} \int_{0}^{\infty}\left[u^{2}(r)+v^{2}(r)\right] r^{2} d r$$ the rms of the deuteron, but I'm in the same place, i don't understand where it comes from the 1/4 factor and how to use it in the $$\widehat{Q}_{0}$$ to obtain $$\widehat{Q}_{d}$$.