I need help with a problem. Problem: Recalling that the Fermic-Dirac distribution function applies to all fermions, including protons and neutrons, each of which have spin 1/2, consider a nucleus of 22Ne consisting of 10 protons and 12 neutrons. Protons are distinguishable from neutrons, so two of each particle (spin up, spin down) can be put into each energy state. Assuming that the radius of the 22Ne nucleus is 3.1*10^-15 m, estimate the fermi energy and the average energy of the nucleus in 22Ne. Express your results in MeV. Do the results seem reasonable? My solution: We are going to estimate the fermi energy Ef. To do that i have integrate n(E)dE from 0 to Ef -> INT[0,Ef](n(E)dE)=INT[0,Ef](g(E)dE) here i use g(E)=(1/h^3)*4*PI*(2*m)^(3/2)*V*E^(1/2). After integration i get -> N=4*PI*(2*m)^(3/2)*V*Ef^(1/2). I solve Ef and get -> Ef=(N/V)^(2/3) * (1/((4*PI)^(2/3)*2*m)) *h^2 for protons i get m=mass=1.67*10^-27 kg N=number of protons in the atom=10 V=volume of the atom= (4/3)*PI*r^3 r=radius of atom=3.1*10^-15 m h=plancks constant=6.63*10^-34 Js Inserting this i get 22 MeV which i wrong the right answers seems to be 516 MeV Where i'm doing wrong?