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# soltn12 - Homework 12 Solutions 1110(a(b m AS...

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1110 (a) Homework 12 Solutions X = polv3 = t.Zxr0-E AS m.When A=12, R=2-Zrxl0-E m. (c) (b) - k(z-rY' t = ------------7-- = r (lx1oe rvn'/c2[trxro-tt cl' lz-t1 .WhenZ=6and R=275x10-6 m,F=152N. (b) ffi=n=n'"ru=j (os x to{ c[3.2 x rot0 decaye/s ci) . _ . ffi=1'E5x1o'" nudei u=Elz=k(z _tyz _ (gxtot w-2/czXrrxto{e c)z(z-r) ,when Z= 5 and R =z75x10-6 m, U =419x10-t3 | =2.62MeY. (d) A=n&; Z=yz, R =7.44x10-rs m, F=379N and U=282x10-u l=l7.6Mev rr72 E, =l2smr"+frn^- n(*re[ror.s Mev/u) = 492 Mev l}l4 (a) rT2r (a) (b) rT22 (a) * =# ="r lvieV/nucleon; agrees wirh Figur,e 13.10. For {O we have, uaing Equation 13.4 E, =ft8xr.flr782s)+(rm8 66s)-05.003 o65I u(9s1.s Mev/u) = 111.e5 v"v For f;N we have E5 =l/1l(7.w2 ffi) + 8(1.008 66s) - 05.0m 10e)l u(e31.5 Mev/u) = u5.49 Mev Thercforc Al, = 3.53 1,1sY. Uee Equation 13.4 to ma for !if.fa; ft= 8.11 MeV/nucteon andfor flMg; F. ? = 7.\$ Mev/nucleon. The binding enerry per nudeon is greater for ffNa by 0210 MeV. In boar case6, th€ iroba\$ with more protone oqrerien€e more couhrnb rep'uldon and ar,e less tighdy bormd. write Equation t3.toar f =c-r, eo,t ", r=1h(&). n t*o* f;=s *rr"r, 1 | =2 h, n 2 = --:-ln 5 = 0.805 h-r.

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soltn12 - Homework 12 Solutions 1110(a(b m AS...

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