I. Solutions to Word Problems on Nuclear Activities

= (1/10) x 10^-6 Ci x 3.7 x 10¹⁰ Bq/Ci = 3.7 x 10³ Bq = 3.7E10.

Solve first for the decay constant (l ) for Polonium using its half life in the formula: R = R_oe^{-l T1/2} , where R=1/2 and R_o=1:

0.5 = e^{-l (1.19E7 sec.)} (Divide both sides by 1/2 = 0.5)

(Ln 0.5)/1.19E7 = -l x Ln e (Take the natural log of both sides)

- 6.93E-1/1.19E7 = -l (Ln e = 1. Check it in your calculator)

l = 5.82E-8/sec = 5.82E-8 (The decay constant for Polonium).

Now, use again the formula R = R_oe^{-l t} by plugging in all known values.

3.67 Bq = 3.7E3 Bq x e ^–5.82E-8t

3.67/3.7E3 = e ^-5.82E-8t

Ln(9.92E-4) = - 5.82E-8t

t = -6.92/-5.82E-8 = 1.19E8 seconds/(60x60x24x365) » 3.77 years

Note: The solution was worked out in details to show how it was done

without taking into consideration the efficiency of the Geiger Tube,

which is not given by the manufacturer.

I. Solutions to Word Problems on Nuclear Activities

= 1x 10^-6 Ci x 3.7 x 10¹⁰ Bq/Ci = 3.70 x 10⁴ Bq = 3.70E4

T_1/2 = 3.8 years x 365 x 24 x 60 x 60 = 1.20 x 10⁸ sec = 1.20E8

Solve for the decay constant (l ) for Thallium using its half life in the formula: R = R_oe^{-l T1/2} , where R=1/2 and R_o=1:

1/2 = 1 e^{-l (1.20E8 sec.)} (Use half-life when 1 decays to 1/2).

0.5/1 = e^{-l (1.20E8 sec.)} (1/2 = 0.5)

Ln .5 = -l (1.20E8 sec.) x Ln(e) (Take the natural log of both sides).

-6.93E-1/1.20E8 = -l (Ln e = 1)

l = -5.78E-9/sec (The decay constant for Thallium).

Now, use again the formula R = R_oe^{-l t} by plugging in all known values.

205 Bq = 3.7 x 10⁴ Bq x e ^–5.78E-9t

205/3.7E4 = e ^–5.78E-9t

Ln(5.54E-3) = - 5.78E-9t

t = -5.20/-5.78E-7 = 8.99E8 sec/(60x60x24x365) » 28.5 years

(Approximate only because the efficiency of the Geiger Tube is not given.)

I. Solutions to Word Problems on Nuclear Activities

half-life, T_1/2 = 5.3 years

= 1x 10^-6 Ci x 3.7 x 10¹⁰ Bq/Ci = 3.70 x 10⁴ Bq. = 3.70E4

T_1/2 = 5.3 years x 365 x 24 x 60 x 60 = 1.67 x 10⁸ sec = 1.67E8

Solve for the decay constant (l ) for Cobalt using its half life in the formula: R = R_oe^{-l T1/2} , where R=1/2 and R_o=1:

1/2 = 1 e^{-l (1.67E8 sec.)} (Use half life when 1 decays to 1/2)

0.5/1 = e^{-l (1.67E8 sec.)} ( ½=0.5)

Ln 0.5 = -l (1.67E8 sec.) x Ln(e) (Take the natural log of both sides).

-6.93E-1/1.67E8 = -l (Ln e = 1)

l = 4.15E-9/sec = 4.15E-9/sec (The decay constant for Cobalt).

Now, use again the formula R = R_oe^{-l t} by plugging in all known values.

85.3 Bq = 3.7E4 Bq x e ^–4.15E-9t

85.3/3.7E4 = e ^–4.15E-9t

Ln(2.31E-3) = - 4.15E-9t

t = -6.07/-4.15E-9 = 1.46E9 seconds/(60x60x24x365) » 46.40 years

(Approximate only because the efficiency of the Geiger Tube is not given.)

II. Solutions to Word Problems on Nuclear Activities

Find m _o, the initial mass of each source at the time it was manufacture.

1 u (atomic mass unit) = 1.67265E-24 g

1 mol (mole) = 6.002E23 u = 1 g of matter

1 Ci = 3.7E10 Bq (decays/sec)

1. Polonium Po-210 (alpha decay): Activity, R_o = 1/10 m Ci.

Mass of Po-210 atom = 210 u x 1.67264E-24 g/u = 3.51E-22 g/atom

R_o = (1/10)E-6 Ci. x 3.7E10 Bq/Ci = 3.70E3 Bq (decays per second)

N_o = R_o/l = (3.70E3 Bq) / |-5.82E-8/sec| = 6.36E10 atoms initially present

m_o (g ) = 6.36E10 atoms x 3.51E-22 g/atom = 2.23E-11 g.

2. Thallium Tl-204 (beta decay): Activity, R_o = 1 m Ci.

Mass of Tl-204 atom = 204 u x 1.67264E-24 g/u = 3.41E-22 g/atom

R_o = 1E-6 Ci x 3.7E10 Bq/Ci = 3.70E4 Bq (decays per second)

N_o = R_o/l = (3.70E4 Bq) / |-5.78E-9/sec | = 6.40E12 atoms initially present

m_o (g) = 6.40E12 atoms x 3.41E-22 g/atom = 1.07E-11 g.

Mass of Co-60 atom = 60 u x 1.67264-24 g/u = 1.0E-22 g/atom

R_o = 1.0E-6 Ci x 3.7E10 Bq/Ci = 3.70E4 Bq (decays per second)

N_o = R_o/l = (3.70E4 Bq) / |-4.15E-9/sec | =9.02E12 atoms initially present

m_o (g ) = 9.02E12 atoms x 1.0E-22 g/atom = 9.02E-10 g.