Solutions To Basic Radioactivity Problems (From January 11 Post)

 1) 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.

Solution:   The decay curve is shown below:

We see on inspection that the counts per minute decrease from 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

2)  Find the half life of the beta particle emitting nuclide: 

 ³²P ₁₅,

If the activity A = 5.6 x 10^-7 /s.

Solution:

A = 5.6 x 10^-7 /s.

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

(T½) = 1.24 x 10⁶ s = 14.3 days

3) A radionuclide sample of N = 10¹⁵ atoms undergoes decay at the constant average rate of dN/dt =  6.00 x  10¹¹  /s.  

From this information, find:

a) The Activity A

b) The decay constant l

c) The half life of the sample in minutes.

Solutions:

Given:    dN/dt = - lN

So that: the decay constant l   =    1/N | dN/dt |  

 =  10 ^-15 (6.00 x 10¹¹/s)

l   =  6.00 x 10^-4 s^-1

A = - lN   =   - ( 6.00 x 10^-4 s^-1 )( 10 ¹⁵ )=   -  6.00 x 10¹¹/s  i.e.

6.00 x 10¹¹    decays per second

c)  Half life: T_½  =  ln 2/l    = 0.693/ l =

0.693/ (6.00 x  10^-4 s^-1)

 

=  1160 s  (or 19.3 minutes) 

4) The activity of a radio-nuclide is given as:

A =  A_o  exp (-lt)

Where  A_o is the activity (decay rate) at time t = 0, and A refers to the activity at some time t thereafter. If a particular radio-nuclide has  A_o  = 1. 1 x 10 ¹⁰  decays/sec and a half life T_1/2 = 28.0 years, find:

a) The decay constant, l ,

b) The activity  A after 1 hour,  after 2 hours.

c) The activity  A  after 49 years, 

Solutions:

a)The decay constant  l = 0.693/ T_1/2   =

 0.693/ 883612800s

Or: l =  7.84 x 10 ^-10      

b) The activity of a radio-nuclide is given as:

A =  A_o  exp (-lt)

 

Where  A_o is the activity at time t = 0, and A refers to the activity at some time t thereafter.   The activity A  after 1 hour (3600 s) is given by:

A =  A_o  exp (-lt)  = 1. 1 x 10 ¹⁰  decays/sec (exp (-lt) )

Where (lt)  =  (7.84 x 10 ^{- 10} ) (3600 s) = 2.82 x 10 ^{- 6}

Then:

A_o  exp (-lt)  = 1. 1 x 10 ¹⁰  /s [exp (-2.82 x 10 ^{- 6})] =

1. 1 x 10 ¹⁰  /s [1.00000] =  1. 1 x 10 ¹⁰  decays/sec

 

After two hours (t = 7200s ):

(lt)  =  (7.84 x 10 ^{- 10} ) (7200 s) = 5.64 x 10 ^{- 6}

 A_o  exp (-lt)  = 1. 1 x 10 ¹⁰  /s [exp (-5.64 x 10 ^{- 6})] =

1. 1 x 10 ¹⁰  /s [1.0000] =  1. 1 x 10 ¹⁰  decays/sec

After 49 years, T = 1. 54 x 10 ⁹  sec

lt  =  (7.84 x 10 ^{- 10} ) (1. 54 x 10 ⁹   s) = 1.20

A_o  exp (-lt)  = 1. 1 x 10 ¹⁰  /s [exp (- 1.20)]

 

 =1. 1 x 10 ¹⁰  /s (0.301) =     3.31 x 10 ⁹   decays/sec