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
N = 2.544 × 1015
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Part (b)
Activity after 2 mins is calculated from the following relation
------------------------------------------------
Part (c)
Radioactivity is given as ( dN/dt ) = λ N
If (dN/dt ) = 1 Bq , then N = 1/λ = T1/2 / ln(2) ≈ 44
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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