diff options
Diffstat (limited to '2939/CH4')
35 files changed, 590 insertions, 0 deletions
diff --git a/2939/CH4/EX4.1/Ex4_1.sce b/2939/CH4/EX4.1/Ex4_1.sce new file mode 100755 index 000000000..4f1cbc4b9 --- /dev/null +++ b/2939/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,12 @@ +
+// Ex4_1
+
+clc;
+
+//Given:
+t1=1600;// in year
+a=11.6*10^17;// atoms
+// Solution:
+k=0.693/t1;// year^-1
+L=(a*226)/k;// atomic mass of Radon is 226
+printf("The value of avagadro constant is = %f atoms per mole",L)
diff --git a/2939/CH4/EX4.10/Ex4_10.sce b/2939/CH4/EX4.10/Ex4_10.sce new file mode 100755 index 000000000..3ba12a698 --- /dev/null +++ b/2939/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,15 @@ +
+// Ex4_10
+clc;
+
+// Given
+E=2.5;// in MeV
+// Solution:
+k=0.693/(5.27*3.16*10^7);// decay constant
+A=k*0.1*6.022*10^23;// atoms/s
+A1=3.6*10^3*A;// atoms /hr
+
+E1=A1*E*1.6*10^-13*10^-3;//Energy in KJ/hr
+
+printf("The total energy dissipate per hour is = %f KJ",E1)
+
diff --git a/2939/CH4/EX4.11/Ex4_11.sce b/2939/CH4/EX4.11/Ex4_11.sce new file mode 100755 index 000000000..9695efcf3 --- /dev/null +++ b/2939/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,17 @@ +
+// Ex4_11
+clc;
+
+// Given:
+Ma=4;//mass of alpha particle
+Mr=228;// mass of Th
+Ea=4; //in MeV
+
+
+// Solution:
+Er=(Ma/Mr)*Ea;// energy of recoil
+Et=Ea+Er;// total energy of transition
+
+dM=Et/931;// net mass loss in u
+
+printf("The net mass loss is = %f u",dM)
diff --git a/2939/CH4/EX4.12/Ex4_12.sce b/2939/CH4/EX4.12/Ex4_12.sce new file mode 100755 index 000000000..c108b0c4e --- /dev/null +++ b/2939/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,23 @@ +
+// Ex4_12
+clc;
+
+// Given:
+Ma=4;//mass of alpha particle
+Mr1=222;// mass of
+Mr2=208;
+Ea1=4.863;
+Ea2=6.082;
+// Solution:
+
+Er1=(Ma/Mr1)*Ea1;
+Et1=Ea1+Er1;
+printf("For Ra emitting alpha")
+printf("\n\tEnergy of recoil is %f MeV",Er1)
+printf("\n\tTotal transition energy is %f MeV",Et1)
+Er2=(Ma/Mr2)*Ea2;
+Et2=Ea2+Er2;
+printf("\nFor Bi emitting alpha")
+printf("\n\tEnergy of recoil is %f MeV",Er2)
+printf("\n\tTotal transition energy is %f MeV",Et2)
+
diff --git a/2939/CH4/EX4.13/Ex4_13.sce b/2939/CH4/EX4.13/Ex4_13.sce new file mode 100755 index 000000000..aec61ea44 --- /dev/null +++ b/2939/CH4/EX4.13/Ex4_13.sce @@ -0,0 +1,16 @@ +
+// Ex4_13
+clc;
+
+// Given:
+dm=0.006332;// in u
+ma=4;
+mCm=244;
+
+// Solution:
+
+E=dm*931;// in MeV
+KE=E*(ma/mCm); // in MeV
+v=sqrt((2*KE*1.6*10^-13)/(240*1.6605*10^-27));
+printf("The Kinetic Energy and velocity are %f MeV and %f m/s respectively",KE,v)
+
diff --git a/2939/CH4/EX4.14/Ex4_14.sce b/2939/CH4/EX4.14/Ex4_14.sce new file mode 100755 index 000000000..1898ca4b2 --- /dev/null +++ b/2939/CH4/EX4.14/Ex4_14.sce @@ -0,0 +1,14 @@ +
+// Ex4_14
+clc;
+
+// Given:
+E0=1.7;// in MeV
+// Solution:
+// For E0<2.5 MeV; using Katz and Penfold empirical equation we have
+R1=412*((E0)^(1.265-0.0954*log(E0)));// mg/cm^2
+// Using feather's relation we have
+R2=530*E0-106;// mg/cm^2
+
+printf("The range in Al for beta radiation is %f mg/cm^2 using Katz and Penfold empirical equation and %f mg/cm^2 using feathers relation.",R1,R2)
+
diff --git a/2939/CH4/EX4.15/Ex4_15.sce b/2939/CH4/EX4.15/Ex4_15.sce new file mode 100755 index 000000000..d9c555538 --- /dev/null +++ b/2939/CH4/EX4.15/Ex4_15.sce @@ -0,0 +1,13 @@ +
+// Ex4_15
+clc;
+
+
+// Solution:
+L1=(5.5-3.5);// Case 1
+L2=2-0;// Case 2
+L3=1.5-.5;// Case 3
+ELML1=1+0+2;
+ELML2=1+1+2;
+ELML3=0+1+1;
+printf("The order for Case 1 is %f and emission is type (%f) M2,Case 2 is %f and emission is type (%f) E2,Case 3 is %f and emission is type (%f) E1,Case 4 is not possible.",L1,ELML1,L2,ELML2,L3,ELML3)
diff --git a/2939/CH4/EX4.16/Ex4_16.sce b/2939/CH4/EX4.16/Ex4_16.sce new file mode 100755 index 000000000..1777c7440 --- /dev/null +++ b/2939/CH4/EX4.16/Ex4_16.sce @@ -0,0 +1,19 @@ +
+// Ex4_16
+clc;
+
+// Given:
+m=4*10^-3;// in gms
+M=210;
+E=0.34;// in MeV
+// Solution:
+
+N=(m*6.022*10^23)/M;
+k=0.693/(5*24*3600);// in s^-1
+A=N*k;// in dis/s
+// Energy released at 0.34 MeV per dis/s will be
+E1=E*A;// in MeV/s
+E2=E1*1.6*10^-13;// watts
+
+printf("The rate of energy emission is %f W",E2)
+
diff --git a/2939/CH4/EX4.17/Ex4_17.sce b/2939/CH4/EX4.17/Ex4_17.sce new file mode 100755 index 000000000..8a7e65203 --- /dev/null +++ b/2939/CH4/EX4.17/Ex4_17.sce @@ -0,0 +1,10 @@ +
+// Ex4_17
+clc;
+// Given:
+A=0.2506*10^15;// atoms/s re: Ex4_10
+// Solution:
+Strength=A/(3.7*10^10);// in kCi
+S1=Strength*10^-3;// in KCi
+printf("The strength in KCi is %f ",S1)
+
diff --git a/2939/CH4/EX4.18/Ex4_18.sce b/2939/CH4/EX4.18/Ex4_18.sce new file mode 100755 index 000000000..85b47dca0 --- /dev/null +++ b/2939/CH4/EX4.18/Ex4_18.sce @@ -0,0 +1,24 @@ +
+// Ex4_18
+clc;
+
+// Given:
+N1=10^24;// atoms
+N2=10^16;// atoms
+N3=1000;// atoms
+N4=80;// atoms
+
+// Solution:
+N11=N1*0.5;// 1st half life
+N12=N11/2;// 2nd half life
+N13=N12/2;// 3rd half life
+printf("\n The 1st half life, 2nd half life, 3rd half life are %f, %f, %f respectively.",N11,N12,N13)
+N21=N2/2;// 1st half life
+N22=N21/2;// 2nd half life
+N23=N22/2;// 3rd half life
+printf("\n The 1st half life, 2nd half life, 3rd half life are %f, %f, %f respectively.",N21,N22,N23)
+N31=N3/2;// 1st half life
+N32=N31/2;// 2nd half life
+N33=N32/2;// 3rd half life
+printf("\n The 1st half life, 2nd half life, 3rd half life are %f, %f, %f respectively.",N31,N32,N33)
+//Radiactivity is a statistical property. Decay kinetics are reliable only when initial number is large
diff --git a/2939/CH4/EX4.2/Ex4_2.sce b/2939/CH4/EX4.2/Ex4_2.sce new file mode 100755 index 000000000..6f713ff17 --- /dev/null +++ b/2939/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,13 @@ +
+// Ex4_2
+clc;
+
+// Given:
+t1=1.3*10^9;// in years
+w=0.0119;// wt %
+
+// Solution:
+N=(w*6.022*10^23)/(40*100);
+k=(0.693*60)/(t1*3.16*10^7);
+sa=N*k;// specific activity
+printf("The specific activity is = %f dis min^-1 g^-1",sa)
diff --git a/2939/CH4/EX4.20/Ex4_20.sce b/2939/CH4/EX4.20/Ex4_20.sce new file mode 100755 index 000000000..95dbf7764 --- /dev/null +++ b/2939/CH4/EX4.20/Ex4_20.sce @@ -0,0 +1,16 @@ +
+// Ex4_20
+clc;
+
+// Given:
+t1=1.28*10^9;// in years
+// Solution:
+k=0.693/(1.28*10^9);
+// beta deay is 88.8%
+k1=0.888*k;
+// EC decay is 11.2%
+k2=0.112*k;
+tbeta=(0.693*10^-9)/(k1);// partial half life for beta decay in Gy
+tEC=(0.693*10^-9)/(k2);// partial half life for EC decay in Gy
+
+printf("The partial half life for beta decay is %f Gy and partial half life for EC decay is %f Gy.",tbeta,tEC)
diff --git a/2939/CH4/EX4.21/Ex4_21.sce b/2939/CH4/EX4.21/Ex4_21.sce new file mode 100755 index 000000000..c2963c34f --- /dev/null +++ b/2939/CH4/EX4.21/Ex4_21.sce @@ -0,0 +1,12 @@ +
+// Ex4_21
+clc;
+// Given:
+t=15.02;// in hours
+// Solution:
+ar=1000;// activity ratio given that 0.1% of intial activity
+k=0.693/t;
+
+t1=(log(ar))/k;
+printf("The time required will be %f h",t1)
+
diff --git a/2939/CH4/EX4.22/Ex4_22.sce b/2939/CH4/EX4.22/Ex4_22.sce new file mode 100755 index 000000000..86a3ab947 --- /dev/null +++ b/2939/CH4/EX4.22/Ex4_22.sce @@ -0,0 +1,14 @@ +
+// Ex4_22
+clc;
+// Given:
+t=6.01;// in hours
+// Solution:
+ar=100/5;// activity ratio given that 5% of intial activity
+k=0.693/t;
+
+t1=(log(ar))/k;
+printf("The time required will be %f h",t1)
+
+
+
diff --git a/2939/CH4/EX4.23/Ex4_23.sce b/2939/CH4/EX4.23/Ex4_23.sce new file mode 100755 index 000000000..9bbbf28b8 --- /dev/null +++ b/2939/CH4/EX4.23/Ex4_23.sce @@ -0,0 +1,23 @@ +
+//Ex4_23
+clc;
+
+// Given:
+
+t=1.83*10^9;// in years
+// Solution:
+// Part (a)
+k=(0.693)/(t*3.16*10^7);
+k1=(0.693*10^17)/(t*3.16*10^7);// in 10^-17 s^-1
+printf("\n The overall decay constant will be %f*10^-17 s^-1",k1)
+// Part (b)
+a=(6.022*10^23)/40; // atoms of K(40)
+A=a*k;// activity
+printf("\n The activity for k(40) is %f beta/s",A)
+
+// Part (c)
+a1=(6.022*10^23*1.2*10^-4)/41; // atoms of K(41)
+A1=a1*k;// activity
+printf("\n The activity for k(41) is %f beta/s",A1)
+
+
diff --git a/2939/CH4/EX4.24/Ex4_24.sce b/2939/CH4/EX4.24/Ex4_24.sce new file mode 100755 index 000000000..2128bde0e --- /dev/null +++ b/2939/CH4/EX4.24/Ex4_24.sce @@ -0,0 +1,12 @@ +
+// Ex4_24
+clc;
+
+// Given:
+a1=6520;// c/min
+a2=4820;//c/min
+t=2;//min
+// Solution:
+k=log(a1/a2)/t;
+t1=0.693/k;// half life
+printf("The decay constant is %f min^-1 and the half life is %f min",k,t1)
diff --git a/2939/CH4/EX4.25/Ex4_25.sce b/2939/CH4/EX4.25/Ex4_25.sce new file mode 100755 index 000000000..f659cbe93 --- /dev/null +++ b/2939/CH4/EX4.25/Ex4_25.sce @@ -0,0 +1,12 @@ +
+// Ex4_25
+clc;
+// Given:
+a=(1/32);// activity drop of its initial value
+t1=7.5;//in h case(a)
+t2=64.45;// in min case(b)
+// Solution:
+n=log(a)/log(0.5);
+t11=t1/n;// half life
+t12=t2/n;// half life
+printf("The half life for case (a)is %f h and case(b) is %f min",t11,t12)
diff --git a/2939/CH4/EX4.26/Ex4_26.sce b/2939/CH4/EX4.26/Ex4_26.sce new file mode 100755 index 000000000..3efcf7bff --- /dev/null +++ b/2939/CH4/EX4.26/Ex4_26.sce @@ -0,0 +1,23 @@ +
+// Ex4_26
+clc;
+
+// Given:
+t238=4.5*10^9;// in y
+t235=7.04*10^8;// in y
+a0=0.72;// atoms per cent
+t=7.04*10^8;
+// Solution:
+
+k1=0.693/(t238);//decay constant for U 238
+N1=(100-a0)*exp(k1*t);
+
+k2=0.693/(t235);//decay constant for U 235
+N2=(a0)*exp(k2*t);
+
+proportion=N2/N1;
+printf("The proportion of U235 704 million years back is %f" ,proportion)
+
+
+
+
diff --git a/2939/CH4/EX4.27/Ex4_27.sce b/2939/CH4/EX4.27/Ex4_27.sce new file mode 100755 index 000000000..212a79297 --- /dev/null +++ b/2939/CH4/EX4.27/Ex4_27.sce @@ -0,0 +1,13 @@ +
+// Ex4_27
+clc;
+
+// Given:
+t=110;// in min
+a=10;//dpmg^-1
+// Solution:
+k=0.693/t;
+N=a/k;// atoms of F18
+mass=(N*18)/6.022*10^23;
+mass1=(N*18*10^20)/(6.022*10^23);// in 10^-20 grams
+printf("The no. of atoms produced is %f and its mass is %f*10^-20 grams",N,mass1)
diff --git a/2939/CH4/EX4.28/Ex4_28.sce b/2939/CH4/EX4.28/Ex4_28.sce new file mode 100755 index 000000000..461bab776 --- /dev/null +++ b/2939/CH4/EX4.28/Ex4_28.sce @@ -0,0 +1,15 @@ +
+// Ex4_28
+clc;
+
+// Given:
+t=14.3;// half life in days
+// Solution:
+k=0.693/(t*24*3600);
+N=(3.7*10^10)/(k);// No. of atoms in 1 Ci
+N1=N*(1-(exp(-0.693/14.3)));// atoms of S32 produced
+mass=(N1*32)/(6.022*10^23);
+
+m1=mass*10^6;// in micro grams
+printf("The mass in micro-grams is %f.",m1)
+
diff --git a/2939/CH4/EX4.29/Ex4_29.sce b/2939/CH4/EX4.29/Ex4_29.sce new file mode 100755 index 000000000..c1e70d45a --- /dev/null +++ b/2939/CH4/EX4.29/Ex4_29.sce @@ -0,0 +1,26 @@ +
+// Ex4_29
+clc;
+
+// Given:
+t=3.82;// in days
+// Solution:
+// part(a)
+days=1;
+D1=(1-(exp(-0.693*days/t)))*100;
+printf("\n The fraction decayed in 1 day will be %f.",D1)
+// part(b)
+days=5;
+D1=(1-(exp(-0.693*days/t)))*100;
+printf("\n The fraction decayed in 5 days will be %f.",D1)
+// part(c)
+days=10;
+D1=(1-(exp(-0.693*days/t)))*100;
+printf("\n The fraction decayed in 10 days will be %f.",D1)
+// part(d)
+days=6*t;
+D1=(1-(exp(-0.693*days/t)))*100;
+printf("\n The fraction decayed in 6 half lives will be %f.",D1)
+// part(e)
+n=log(0.001)/log(0.5);
+printf("\n Time needed for the decay of 99.9 percent is %f half lives i.e.%f days.",n,t*n)
diff --git a/2939/CH4/EX4.3/Ex4_3.sce b/2939/CH4/EX4.3/Ex4_3.sce new file mode 100755 index 000000000..7cba1711b --- /dev/null +++ b/2939/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,24 @@ +
+// Ex4_3
+
+clc;
+
+// Given:
+L=6.022*10^23;
+// Solution:
+// 1 mCi= 3.7*10^7 dis/s
+k1=0.693/(15*3600);
+N1=3.7*10^7/k1;
+m1=(24*N1*10^10)/L;
+printf("\n The no. of atoms of Na(24) are = %f",N1)
+printf("\n The mass of Na(24) is %f * 10^-10 g",m1)
+k2=0.693/(14.3*24*3600);
+N2=3.7*10^7/k2;
+m2=(32*N2*10^9)/L;
+printf("\n \n The no. of atoms of P(32) are = %f",N2)
+printf("\n The mass of P(32) is %f * 10^-9 g",m2)
+k3=0.693/(1600*3.16*10^7);
+N3=3.7*10^7/k3;
+m3=(226*N3*10^3)/L;
+printf("\n \n The no. of atoms of Ra(226) are = %f",N3)
+printf("\n The mass of Ra(226) is %f * 10^-3 g",m3)
diff --git a/2939/CH4/EX4.30/Ex4_30.sce b/2939/CH4/EX4.30/Ex4_30.sce new file mode 100755 index 000000000..6c0a1af03 --- /dev/null +++ b/2939/CH4/EX4.30/Ex4_30.sce @@ -0,0 +1,17 @@ +
+// Ex4_30
+clc;
+// Given:
+t=2.6;// years
+// Solution:
+k=0.693/t;// decay constant
+//part(a)
+kbeta=0.89*k;
+printf("\n The decay constant is %f y^-1",kbeta)
+kEC=0.11*k;
+printf("\n The decay constant is %f y^-1",kEC)
+//part(b)
+tbeta=0.693/kbeta;
+printf("\n The half life is %f y",tbeta)
+tEC=0.693/kEC;
+printf("\n The half life is %f y",tEC)
diff --git a/2939/CH4/EX4.31/Ex4_31.sce b/2939/CH4/EX4.31/Ex4_31.sce new file mode 100755 index 000000000..2a29947fa --- /dev/null +++ b/2939/CH4/EX4.31/Ex4_31.sce @@ -0,0 +1,27 @@ +
+// Ex4_31
+clc;
+// Given:
+t=12.8;// hours
+// Solution:
+k=0.693/t;// decay constant
+//part(a)by EC
+kEC=0.42*k;
+printf("Decay by EC")
+printf("\n\t The decay constant is %f h^-1",kEC)
+tEC=0.693/kEC;
+printf("\n\t The half life is %f h",tEC)
+
+//part(b)by beta+
+kbeta1=0.19*k;
+printf("\nDecay by beta+")
+printf("\n\t The decay constant is %f h^-1",kbeta1)
+tbeta1=0.693/kbeta1;
+printf("\n\t The half life is %f h",tbeta1)
+
+//part(b)by beta+
+kbeta2=0.39*k;
+printf("\nDecay by beta-")
+printf("\n\t The decay constant is %f h^-1",kbeta2)
+tbeta2=0.693/kbeta2;
+printf("\n\t The half life is %f h",tbeta2)
diff --git a/2939/CH4/EX4.32/Ex4_32.sce b/2939/CH4/EX4.32/Ex4_32.sce new file mode 100755 index 000000000..feafead94 --- /dev/null +++ b/2939/CH4/EX4.32/Ex4_32.sce @@ -0,0 +1,14 @@ +
+//Ex4_32
+clc;
+
+// Given:
+tU=4.47*10^9;// y
+tRa=1600;// y
+tRn=3.82;// days
+nU=1;
+// Solution:
+//under secular equilibrium we have
+nRa=(tRa*365/tRn)*nU;
+nRn=(tU*365/tRn)*nU;
+printf("The proportion of U,Ra,Rn is 1: %f: %f .",nRa,nRn)
diff --git a/2939/CH4/EX4.33/Ex4_33.sce b/2939/CH4/EX4.33/Ex4_33.sce new file mode 100755 index 000000000..76b3960a9 --- /dev/null +++ b/2939/CH4/EX4.33/Ex4_33.sce @@ -0,0 +1,19 @@ +
+//Ex4_33
+clc;
+
+// Given:
+ax0 =1; //assume
+tx = 2; //hrs
+ty = 1; //hrs
+// Solution:
+// general equation connecting Ax and Ay is
+// Ax(n) = (ky * Ax(0) * (exp(-kx * t) - exp(-ky * t))/ (ky - kx)) + Ay(0) * exp(-ky * t)
+ax0 = 1;
+ay4 = (ax0 * (0.693/1) * ((1/4)-(1/16)))/((0.693/1)-(0.693/2)) + ax0 * (1/16);
+
+ax4 = (1* ax0)/4;
+
+proportion = (ay4 * 100)/(ay4 + ax4);
+
+printf("The proportion of ay4 at the end of 4 hrs is %f",proportion)
diff --git a/2939/CH4/EX4.34/Ex4_34.sce b/2939/CH4/EX4.34/Ex4_34.sce new file mode 100755 index 000000000..504684b34 --- /dev/null +++ b/2939/CH4/EX4.34/Ex4_34.sce @@ -0,0 +1,20 @@ +// Ex4_34
+clc;
+
+//Given:
+Ax0 = 2000; //dps
+//Solution:
+//part a
+ky = 0.693/10;
+kx = 0.693/288;
+// general equation connecting Ax and Ay is
+ Ax12 = (ky * Ax0 * (0.5^(1/24) - 0.5^(1.2)))/ (ky - kx);
+
+printf("\n Activity due to La(140) at the end of 12 hrs will be %f dps",Ax12);
+//part b
+ky = 0.693/10;
+kx = 0.693/288;
+// general equation connecting Ax and Ay is
+ Ax24 = (ky * Ax0 * (0.5^(2) - 0.5^(57.6)))/ (ky - kx);
+
+printf("\n Activity due to La(140) at the end of 24 d will be %f dps",Ax24);
diff --git a/2939/CH4/EX4.35/Ex4_35.sce b/2939/CH4/EX4.35/Ex4_35.sce new file mode 100755 index 000000000..2253c2706 --- /dev/null +++ b/2939/CH4/EX4.35/Ex4_35.sce @@ -0,0 +1,14 @@ +
+// Ex4_35
+clc;
+
+// Given:
+
+t1=2.7;// h
+t2=3.6;// h
+// Solution:
+k1=0.693/t1;
+k2=.693/t2;
+
+tmax=(log(k2/k1))/(k2-k1);
+printf("The time when daughter activity reaches maximum is %f and this is same when activities of both are equal.",tmax)
diff --git a/2939/CH4/EX4.36/Ex4_36.sce b/2939/CH4/EX4.36/Ex4_36.sce new file mode 100755 index 000000000..cea45b2be --- /dev/null +++ b/2939/CH4/EX4.36/Ex4_36.sce @@ -0,0 +1,13 @@ +
+//Ex4_36
+clc;
+// Given:
+tPo=138;// days
+n=24.86;// days
+// Solution:
+kPo = 0.693/tPo;
+// using simplification logx=2(x-1)/(x+1)
+kBi=((2 * 2.303)-(n*kPo))/n;
+tBi=0.693/kBi;
+
+printf("The half life of Bi is %f days",tBi)
diff --git a/2939/CH4/EX4.37/Ex4_37.sce b/2939/CH4/EX4.37/Ex4_37.sce new file mode 100755 index 000000000..a55d89397 --- /dev/null +++ b/2939/CH4/EX4.37/Ex4_37.sce @@ -0,0 +1,13 @@ +
+//Ex4_37
+clc;
+
+// Given:
+a=10*10^7;// rate
+t=15;// h
+// Soution:
+A30=a*(1-(0.5)^(2));// dps
+
+A45=A30*((0.5)^(3));// dps
+
+printf("The residual activity in the sample is %f dps",A45)
diff --git a/2939/CH4/EX4.4/Ex4_4.sce b/2939/CH4/EX4.4/Ex4_4.sce new file mode 100755 index 000000000..6d5439383 --- /dev/null +++ b/2939/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,14 @@ +
+//Ex4_4
+
+clc;
+
+// Given:
+t1=12.3;// in yrs
+L=6.022*10^23;
+// Solution:
+k=.693/(t1*3.16*10^7);// in s^-1
+A=(2*L)/(2.24*10^4);// no. of atoms
+a1=A*k;// dis per s
+a=a1/(3.7*10^10);// activity in Ci/cm^3
+printf("The activity in Ci/cm^3 = %f",a)
diff --git a/2939/CH4/EX4.5/Ex4_5.sce b/2939/CH4/EX4.5/Ex4_5.sce new file mode 100755 index 000000000..88ef40aaf --- /dev/null +++ b/2939/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,17 @@ +
+// Ex4_5
+
+clc;
+
+// Given:
+t=5736;// in years
+Nk=16.1;// dis/min
+L=6.022*10^23;
+// Solution:
+k=(0.693*60)/(t*3.16*10^7);
+N1=Nk/k;// atoms per g for C14
+N2=L/12;//
+
+r=(N1*10^12)/N2;// ratio of C14/C12 in atmosphere
+
+printf("The ratio of C14/C12 in atmosphere in 10^-12 is = %f",r)
diff --git a/2939/CH4/EX4.7/Ex4_7.sce b/2939/CH4/EX4.7/Ex4_7.sce new file mode 100755 index 000000000..ec33900fb --- /dev/null +++ b/2939/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,19 @@ +
+//Ex4_7
+clc;
+
+// Given:
+dA = 206-238;
+dA_Beta=0;
+dA_Alpha = -4;
+
+dZ_Alpha = -2;
+dZ_Beta = 1;
+nBeta=0; //random initialisation
+dZ = 82 -92;
+// Solution:
+nAlpha = (dA- (dA_Beta* nBeta))/dA_Alpha;
+
+nBeta = (dZ- (dZ_Alpha * nAlpha))/dZ_Beta;
+
+printf("Number of alpha decays =%f and number of beta decays = %f",nAlpha,nBeta);
diff --git a/2939/CH4/EX4.8/Ex4_8.sce b/2939/CH4/EX4.8/Ex4_8.sce new file mode 100755 index 000000000..1f653b30e --- /dev/null +++ b/2939/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,26 @@ +// Ex4_8
+
+clc;
+// Given:
+E1=0.059;
+E2=2.5;
+E3=1.33;
+Ei=0;
+Ef=0;
+
+// Solution:
+// delta E for 1,2 & 3 photon
+dE1=E1-Ei;
+dE2=E2-E3;
+dE3=E3-Ef;
+// delta I for 1,2 & 3 photon
+dI1=2-5;
+dI2=4-2;
+dI3=2-0;
+// EL/ML for 1,2 & 3 photon
+ELML1=3+1+1
+ELML2=2+1+1;
+ELML3=2+1+1;
+printf("\n For first photon, dE1=%f MeV, dI1=%f, since EL/ML1=%f & (L+PI+PF) is odd, M3",dE1,dI1,ELML1)
+printf("\n For second photon, dE2=%f MeV, dI2=%f,since EL/ML2=%f & (L+PI+PF) is even, E2",dE2,dI2,ELML2)
+printf("\n For third photon, dE3=%f MeV, d3I=%f, since EL/ML3=%f & (L+PI+PF) is even, E2",dE3,dI3,ELML3)
diff --git a/2939/CH4/EX4.9/Ex4_9.sce b/2939/CH4/EX4.9/Ex4_9.sce new file mode 100755 index 000000000..7140bf838 --- /dev/null +++ b/2939/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,11 @@ +
+// Ex4_9
+clc;
+
+// Given:
+E=2.5; // in MeV
+// Solution:
+// 1 Mev/atom=96.32GJ/mole
+E1=E*96.32// GJ/mole
+E2=0.1*E1;// for 0.1 mole
+printf("The energy that would be released for 0.1 mole of Co will be = %f GJ",E2)
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