Monday, February 15, 2021

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: 


 3215,

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: T½  =  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 exp (-lt)

Where  Ao 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  Ao  = 1. 1 x 10 10  decays/sec and a half life T1/2 = 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/ T1/2   =

 0.693/ 883612800s

Or: l =  7.84 x 10 -10      

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

A =  A exp (-lt)
 
Where  Ao 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 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 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 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 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   


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