# soltn5 - Homework 5 Solutions wq(^ TLr C,sta.c,c closcs a...

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Homework 5 Solutions wq) (^) TLr C.,sta.c. .,c o{ closcs* a+gn"y^,9 6 Jiwt,^. b) I {rxarXi.v,J"V. "-) b l4aV Powld neel l' \ , tb,l lo tot Qt';"t: \f rn -rl t, qTe a", = ooo .p.)ff# = cr *-{##)' = u.* "p. . 323 Ea= Fe" \$e I f From Equation 4.1O rn" rr"u" an*{T/)' ds scaftered at ,10 degrees is dven by Similarly A n at 60 degrees = I.45 cpm A n at 80 degr€es - 0.533 cpm A r at lfi) degrees = 02fl qm /1\ Frorn 4.16 doublintlr|""vL reducee An by a factor of4. Thus Ar at zo a"r,"* = (l|rm cpm) = !5 spm. From rr.16 we find + = #{so, 4u 29, zn =D. Ane,, Zi"Nt" Nc" = number of Cu nudei per unit area = nunber of Cu nudei per unit volume. foil thicknese =ft" v.-,{e:14-ts'', )! =e.n,rco t L\ v^, [rr.r /.-,{@ffff)]r s.ro " roor (+nf or An2 An I :-j-+. Thue the nurnber t--t-' 1,34 xto nm r*, = 13,4 fm f*= lf,lflaV lrXm)( Iv+ctl.n"'r) Sxro-t so A z.o 6 7^o 12e r, 9 4.# {s.n' n ) on{fr)' (H) =,r., *,'.

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ti.b J L4 ) +25 (a) o) The r^dirs = "+ t(4 o.!if is A,u,^ b t qirgr 1i - /qrta\ [tclmF 1 L7EJ^o n"= (.E/r.F n* tilu* ^- )( rr*onu ^"^Xln)' *;uu , {r), [b) LJc .,. .. fmA tGr- vcloci \ us,". .g Boh.'s 1..,a^];aal,ry,- rD^J;], M, qyr= ats 5o ,tr= nh/'r = n[(hclm)/(^r^)faJ. f l}{oev.nn. I r [r) L -Eii- "G,r,-,v ' *l,r^*], =7 (c) ll^, h^{.[,c e^4^t) i! K- l ^." i Ir,^.') tT)'= i (r.,, ",'1v)1{Sff4)' \r'= 0,009.69c- 0) -11". pafc-nl;r.l eL V-
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soltn5 - Homework 5 Solutions wq(^ TLr C,sta.c,c closcs a...

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