PHYSICS PPR4 HELP PLZZZ!!!!

[blueindigo]

Can someone explain me hoe to solve these qs..
Jun03,Q6 cii

Nov04..Q6,c

Jun05 Q7 a and c!!

plzz.i need help urgently, its all abt nuclear physics of how to find out no. of half lives/time!!
i need full explaination as it involves sum maths and im not takin maths!!

esp qs like these....A=Aoe^ln2t/T!!
i dun know how to find out the t here

 

[musa]

june 03 Q6 cii
simple mass of 6.02*10^23 is 90 so what is the mass of 4.67*10^15 atoms

 

[beacon_of_light]

nov04: A = Ao e- ^ ( λt) so A/Ao = 1/10 ... now substitute the rest of the values and find ur answer..
June05: a) in the first 2.6h 1/2 of Fe is formed... in the next 2.6 h 3/4 of Fe is formed... then 7/8 .... go on and plot this on the graph...

c) the ratio shows that 9/10 of Fe is formed so 1/10 of manganese will remain...again apply the same equation... A=Ao e-(λt) ....

 

[usamah2702]

june 2005 q7 part c

mass of iron/mass of maganese=9/1
so the ratio is 9:1 and the total is 10

using the formula N=N0*e^-lamda*t
u already knw lamda frm b(ii)

so when t=0 N0=10 and when t=t(unknown) N=1
so 1=10*e^-(7.4*10^-5)*t
1/10=e^-(7.4*10^-5)*t

ln(1/10)=7.4*10^-5*t
(ln(1/10))/(7.4*10^-5)=t

ull get t in seconds =3116
which is 8.63 hours

 

[blueindigo]

thank you so much!!!!!
very good explaination!!
but, still abt jun03..its actually part c ii..not 2!!
so yeah, if u cud explain me that part, i.e calculate the ratioA/Ao...need detailed explaination!!!