Past exam of the MPhil in nuclear energy course of the University of Cambridge / 2019 / Module NE1 / 2 / c / i / Solution

The fundamental nonnegative mode is ${\varphi }_0{J}_0(2.405r/R)\cos (\pi z/H)$. Put the origin at the centre of the cylinder, so $-H/2\le z\le H/2$. Neglecting extrapolation imposes zero neutron flux at $r=R$ and $z=\pm H/2$. WriteFor an axisymmetric field, separation of variables in cylindrical coordinates givesChoose the axial separation constant so $Z^{\prime \prime }+{\beta }^{2}Z=0$. Midplane symmetry and the lowest positive axial mode give $Z(z)=\cos (\beta z)$, withThe radial equation isThe solution regular on the axis is the zero-order Bessel function of the first kind, $F(r)\propto J_0(\alpha r)$; the independent second-kind solution is singular there. The first radial zero gives $\alpha R=j_{0,1}\simeq 2.405$. Higher zeros would introduce sign changes and do not describe the fundamental nonnegative neutron flux. Normalize by the central value, using $J_0(0)=1$:The printed $2.405$ is a rounded Bessel function zero, rather than an exact root.