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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 | 16 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 | 16 May, 2022, 03:16: PM

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