## Sunday, July 10, 2011

### Solutions for Part 24 (Radioactivity)

We now examine the solutions for Part 24 (Introduction to Basic Physics) on radioactivity. The problems were:

1) Sketch the radioactive decay curve for the practical problem and confirm that the half life is 6 hrs.

2) A point source of gamma radiation has (T½) = 30 mins. The initial count rate recorded by a G-M tube is 360/s. Find the count rate that would be recorded after 4 half lives. Sketch the decay curve and determine the activity, A.

3) At a certain instant, a sample of a radioactive material contains 10^12 atoms. The half life of the material is 30 days.

a) Calculate the no. of disintegrations in the first second.

b) The time elapsed before 10,000 atoms remain.

c) the count rate corresponding to the time in (b).

4) Find the half life of the beta particle emitting nuclide, 32P 15, if the activity A = 5.6 x 10^-7 /s.

The solutions are as follows:

1) The sketch of the decay curve for Problem (1) is shown, and we note it conforms to the half life (T½) = 6h, already computed.

2) The decay curve is shown for Problem 2 (accompanying sketch graph) and we see on inspection that since the counts per minute decreases frmo 360 to 180 in ONE half life (e.g. 1(T½)). The critical aspect to note is that the time is given in units of HALF LIVES not hours! Thus, since 1(T½)= 30 mins. = 1800s then (T½)= 1800 s. This is confirmed from the curve since the 180 counts/min decreases to 90 in 2 half-lives, and this decreases to 45 in 3 half lives and so on.

The activity A = ln 2/(T½) = 0.693/ 1800s = 3.85 x 10^-4 s

3) (a) The activity is: A = 0.693/(T½)

But: (T½) = 30 days x 24 h/day x 60 m/h x 60 s/m = 2,592,000 s

So: A = 0.693/ (2,592,000 s) = 2.67 x 10^-7 /s

So the no. disintegrations/sec for the sample with 10^12 atoms:

= (2.67 x 10^-7 /s) (10^12 ) = 2.67 x 10^5 = 267,000

(b) The decay curve is an exponential function, so we must have:

N = N(o) exp (-At)

where N = 10^4 and N(o) = 10^12

Then:

10^4 = 10^12 exp (-At)

or 10^-8 = exp (-At)

for which we know the value of A and can solve for t:

e.g. -8 = (-A) (t)

t = 8/A = 8/(2.67 x 10^-7/s) = 2.99 x 10^7 s

t = (2.99 x 10^7 s)/ (86400 s/day) = 346.7 days

t = 348 days (approx.) = 11.6 (T½) - call it 12 half lives

c) Then the disintegrations per second must decrease from 267,000 over 12(T½)

The new rate = 65 disintegrations/sec

(4) A = 5.6 x 10^-7 /s.

(T½) = ln 2/ A = 0.693/ (5.6 x 10^-7 /s)

(T½) = 1.24 x 10^6 s = 14.3 days