55Cs124(Z=55,A=124) has a half life of 30.8s.
(a) If we have 7.8 microgram initially, how many nuclei are presents? How many are present 2 min later?
(b) What is the activity at this time?
(c) After how much time will the activity drop to less than about 1 per second?

Asked by arjunsah797 | 16th May, 2022, 02:17: PM

Expert Answer:

Part (a)
 
Number of atoms in w gram = ( N / A ) × w
 
where N = 6.022 × 1023 is Avagadro number and A is atomic weight
 
Number of atoms in  7.8 microgram = ( 6.022 × 1023  / 124 ) × 7.8 × 10-6 = 3.788 × 1016
 
Number of nuclei at present, No = 3.788 × 1016
 
Number N of nuclei after 2 min is determined from radioactivity equation
 
begin mathsize 14px style N space equals space N subscript o space e to the power of negative lambda t end exponent space equals space N subscript o space e to the power of negative fraction numerator left parenthesis ln 2 right parenthesis space t over denominator T subscript 1 divided by 2 end subscript end fraction end exponent end style
begin mathsize 14px style N space equals space 3.788 space cross times 10 to the power of 16 to the power of space end exponent cross times e to the power of negative fraction numerator ln 2 cross times 120 over denominator 30.8 end fraction end exponent end style
N = 2.544 × 1015
 -----------------------------------------------
Part (b)
 
Activity after 2 mins is calculated from the following relation
 
begin mathsize 14px style fraction numerator d N over denominator d t end fraction equals lambda space N space equals space fraction numerator ln left parenthesis 2 right parenthesis space N over denominator T subscript 1 divided by 2 end subscript end fraction space equals space fraction numerator ln left parenthesis 2 right parenthesis cross times 2.544 space cross times 10 to the power of 15 over denominator 30.8 end fraction space equals space 5.725 space cross times 10 to the power of 13 space B q end style
------------------------------------------------
Part (c)
 
Radioactivity is given as ( dN/dt ) = λ N
 
If (dN/dt ) = 1 Bq ,  then N = 1/λ  = T1/2 / ln(2) ≈ 44

begin mathsize 14px style N space equals space N subscript o space e to the power of negative lambda t end exponent space equals space N subscript o space e to the power of negative fraction numerator left parenthesis ln 2 right parenthesis space t over denominator T subscript 1 divided by 2 end subscript end fraction end exponent end style
begin mathsize 14px style ln open parentheses N subscript o over N close parentheses space equals space fraction numerator left parenthesis ln 2 right parenthesis t over denominator T subscript 1 divided by 2 end subscript end fraction end style
t = [ ln( No / N ) × T1/2 ] / (ln2)
 
after substituting values No = 2.544 × 1015 , N = 44 , T1/2 = 30.8 s , we get
 
t = 1408 s = 23 min 28 s

Answered by Thiyagarajan K | 16th May, 2022, 03:16: PM

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