{VERSION 2 3 "IBM INTEL NT" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 11 0 0 0 1 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "R3 Font 0" -1 256 1 {CSTYLE "" -1 -1 "Helvetica" 1 12 128 0 128 1 2 1 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 \+ Font 2" -1 257 1 {CSTYLE "" -1 -1 "Courier" 1 11 0 128 128 1 2 1 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "HATOM" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 489 "In the Bohr theory of the hydrogen atom the electron is in a c ircular orbit of radius rb, 4rb, 9rb,.................., where the rad ius of the first orbit is 52.92 pm. In the quantum theory the electro n is not confined to particular circular orbits; instead there is a ce rtain probability that the electron will be found between distance r a nd r+dr from the nucleus. For the ground state (lowest possible elect ron energy) this probability is given by P(r)dr=4rb^(-3)*r^2*exp(-2r/r b)*dr. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 130 "(a) For a ground state electron, make a plot of the probability f unction P versus the distance r of the electron from the nucleus." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 244 "(b) In o rder to find the most probable electron-nucleus distance r, differenti ate P with respect to r and set equal to zero. How does the most prob able value of r compare with the radius rb of the first Bohr orbit? I s your graph in agreement?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "(c) Calculate the probability of finding the elec tron between r=0 and 2rb ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "(d) Find the radius of a sphere which a ground st ate electron would spend 90% of its time inside" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "Solution:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Par t A" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "Pr obability function." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "P:=4/(rb^3)* r^2*exp(-2*r/rb);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG,$*(%#rbG! \"$%\"rG\"\"#-%$expG6#,$*&F)\"\"\"F'!\"\"!\"#F0\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "Substitute rb into P;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "rb=5292*10^(-2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/ %#rbG#\"%B8\"#D" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "P:=subs( rb=1323/25,P);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG,$*&%\"rG\"\"# -%$expG6#,$F'#!#]\"%B8\"\"\"#\"&+D'\"+n_o:B" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 20 "Plot graph of P vs r" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "plot(P,r=0..250,title=`Probability Function P vs Distance r`);" }} {PARA 13 "" 1 "" {INLPLOT "6&-%'CURVESG6$7`o7$\"\"!F(7$$\"1LL3x1h6o!#; $\"1XUR!=]/A\"!#?7$$\"1nmTN@Ki8!#:$\"1#zn?4Kxv%F/7$$\"1+]7.K[V?F3$\"1c !omt$GV5!#>7$$\"1LL$3FWYs#F3$\"1J'yFl*e2=F;7$$\"1***\\iSmp3%F3$\"1dEZ< !yH'QF;7$$\"1mmmT&)G\\aF3$\"1mSX**G)G_'F;7$$\"1**\\7`p)*>yF3$\"1I@v0w< G7!#=7$$\"1L$ek`o!>5!#9$\"1@vp!>FP7$$\"1+v=n$ycG\"FT$\"12!Q:(fPWFF P7$$\"1nm\"z>)G_:FT$\"1:_+#oJrh$FP7$$\"1nmm\"Hl1#=FT$\"1O2#='G/'\\%FP7 $$\"1nmT&QU!*3#FT$\"1d\"42o%H[`FP7$$\"1+voHR9cBFT$\"1ri;K=8]hFP7$$\"1L $eRZXKi#FT$\"1kBZprd\"*oFP7$$\"1+](oz#)3(GFT$\"1EpB0_o;vFP7$$\"1m;z>,_ =JFT$\"1R*GpmYp2)FP7$$\"1L$eRAM\\P$FT$\"1ll'Rnyge)FP7$$\"1**\\7G$[8j$F T$\"1?&H#HfBA!*FP7$$\"1m;z%*frhTFT$\"1@n+A\")y(p*FP7$$\"1L$3-Q\\gU%FT$ \"1@8h(H3g#**FP7$$\"1**\\ilFQ!p%FT$\"1kT^;bs35!#<7$$\"1$3FWXKj#[FT$\"1 J1)pieX,\"Fbr7$$\"1n\"HK9#Gi\\FT$\"106g)y1)=5Fbr7$$\"1]7.K=B)4&FT$\"1P J&3`P:-\"Fbr7$$\"1ML$3_\"=M_FT$\"17H_I%>G-\"Fbr7$$\"1n\"H23ERN&FT$\"14 OSYM!G-\"Fbr7$$\"1+]iS1ntaFT$\"1'**HtYk<-\"Fbr7$$\"1L3_+_T$f&FT$\"1GRZ z(\\(>5Fbr7$$\"1mmTg(fJr&FT$\"1ZV#>S1o,\"Fbr7$$\"1M$ekynF)fFT$\"1)RI[q ?q+\"Fbr7$$\"1++]7eP_iFT$\"1LCl)*f#G$**FP7$$\"1****\\Pf!Qz'FT$\"1o_lnX Yd&*FP7$$\"1++](=ubJ(FT$\"1\\CYZ7f)4*FP7$$\"1m;zW(*Q*y(FT$\"1j55Wv8C') FP7$$\"1KL$3F-GN)FT$\"16=Hq]\"\\,)FP7$$\"1LLL$e'3I))FT$\"1$=I:S3(yuFP7 $$\"1+]7.;idYFP7$$\"1Le*[.-d>\"F[x$\"1&o#z')pS1UFP7$$\"1n ;/Egw[7F[x$\"14??,&GVv$FP7$$\"1n\"z%*f%)QI\"F[x$\"1fMadtQBLFP7$$\"1+vo za'=N\"F[x$\"13/soy+!)HFP7$$\"1n;zWho.9F[x$\"1i'=>!3TTEFP7$$\"1++]i>Ad 9F[x$\"1qte5IHDBFP7$$\"1+]i:jf4:F[x$\"1H'[Ft9t/#FP7$$\"1+DJ&>r-c\"F[x$ \"1!=%)RN\")e!=FP7$$\"1+]P4q`;;F[x$\"1XuD%\\Zrc\"FP7$$\"1LL$eM%4n;F[x$ \"1')QE8C#oP\"FP7$$\"1++v$4v5s\"F[x$\"1V'3I#Gi'>\"FP7$$\"1m\"zWn*)*p0\"FP7$$\"1++DJiYB=F[x$\"1_/]YY7A\"*F;7$$\"1Lek.Nyt=F[x$ \"1?k))4yQkzF;7$$\"1+Dc^&zj#>F[x$\"1w`dcRL+pF;7$$\"1LL3-=!y(>F[x$\"1D? Ly(e*))fF;7$$\"1+D\"G8O;.#F[x$\"1af\"p%31c^F;7$$\"1nmm\"*\\[$3#F[x$\"1 eY\"=i^wX%F;7$$\"1n;aQz]O@F[x$\"1H4U'pYi$QF;7$$\"1LekG=4*=#F[x$\"1A'*H /Fb,LF;7$$\"1++]i4TPAF[x$\"1Y1QAmDtGF;7$$\"1L$3F9!z#H#F[x$\"1BRC0RYZCF ;7$$\"1nmmT>KUBF[x$\"1%fPdpy#=@F;7$$\"1+DJqJ8&R#F[x$\"1$[V&=?89=F;7$$ \"1+voa-oXCF[x$\"1Dta?*)ei:F;7$$\"$]#F($\"1h]mdruH8F;-%'COLOURG6&%$RGB G$\"#5!\"\"F(F(-%&TITLEG6#%EProbability~Function~P~vs~Distance~rG-%+AX ESLABELSG6$%\"rG%!G-%%VIEWG6$;F(Ff`l%(DEFAULTG" 2 480 350 350 2 0 1 0 2 9 0 4 2 1.000000 45.000000 45.000000 10030 10061 10056 10074 0 0 0 20030 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 238 261 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Part B" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 43 "Find the derivative of P with respect to \+ r." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "der:=diff(P,r);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%$derG,&*&%\"rG\"\"\"-%$expG6#,$F'#!#]\"%B8F(# \"'+]7\"+n_o:B*&F'\"\"#F)F(#!(+]7$\".T#3;ljI" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "Set derivative equal to zero and solve for r." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "sol:=solv e(der=0,r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG6$\"\"!#\"%B8\"# D" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "evalf(sol);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$\"\"!$\"++++#H&!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Part C" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "Find area under P(r) c urve between r=0 and 2*rb." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "rb:=1323/25;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#rbG#\"%B8\"#D" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "P rb:=int(P,r=0..2*rb);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$PrbG,&-%$e xpG6#!\"%!#8\"\"\"F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "eva lf(Prb);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+Wp'*=w!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Part D" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 "Find radius of \+ sphere for which there is a 90% probability of finding the electron in side ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "ans:=int(P,r=0..a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ansG ,**&%\"aG\"\"#-%$expG6#,$F'#!#]\"%B8\"\"\"#!%]7\"(H.v\"*&F'F0F)F0F-F)! \"\"F0F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "fsolve(ans=.9,a );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+hfG39!\"(" }}}}{MARK "0 0 0 " 2 }{VIEWOPTS 1 1 0 1 1 1803 }