| Homework #2, 2008 | ||||
| 1 | A = Ao e -Lt | |||
| e-Lt (decay factor) | ||||
| t (time) | ||||
| Ao (initial radioactive atoms) | ||||
| A (# of atoms presently radioactive) | ||||
| L = 0.693/t1/2 where t=60 days | ||||
| thus -Lt equals 0.693/14.3 days x 60 days | ||||
| e-2.91 = | 0.05447573 | |||
| Ao = A/e-Lt | ||||
| 917839.9284 | cpm | |||
| 1 mC = 2.22x109 cpm | ||||
| 9.17x105/2.22x109 cpm/mC | 0.000413063 | mC | ||
| 1mC 13P weighs 3.5 x 10-9 g | ||||
| 4.13 x 10-4 mC x 3.5x10-9 g/mC = | 1.4455E-12 | g 32P needed | ||
| 2 | lose 50% of initial activity in 20 minutes | |||
| N=No e -(Lt) | ||||
| .50 = Ap e -Lt | ||||
| ln .50 = ln (e -(0.693/t½)(20 min)) | ln 0.50 = | -0.69315 | ||
| -0.693147181 | equals | -(0.693/t½)(20 min) | ln of both sides cancels e | |
| -0.034657359 | equals | equals -0.693/T1/2 | ||
| t½ = | 19.99575327 | |||
|
3 |
A=Ao e-Lt | |||
| L = 0.693/14.3 t = 36 days | ||||
| A= Ao e-1.744 | ||||
| equals 5mC/5ml x e-1.744 | ||||
| exp(-1.744) = | 0.174819721 | |||
| x=0.1747 mC/ml | x=174.7 uC/ml | |||
|
4 |
Lambda = 0.693/250 days | * 1day/24 hrs * 1hr/60min | ||
| this equals 1.9 * 10-6 min-1 | ||||
| A* = 3.7*107 dps/1mC * 60 sec/1min = 2.22*109 dpm/1mCu | ||||
| N = A*/lambda | ||||
| equals 2.22 * 109 dpm/1mCu/1.9 * 10-6 min-1 | * 65 g Zn/6.025*1023 atoms | |||
| this equals 1.27 x 10-7g 65Zn in 1 mCu | ||||