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9. How many disintegrations occur in the first second from 1.00 mol of a radioactive

nuclide with each of the following half-lives:

(a) 12,000 years

(b) 12 hours

(c) 12 minutes

(d) 12 seconds​

Answer :

Answer:

For 12000 years it is 1.1 × 10¹²

For 12 hours it is 9.66 × 10¹⁸

For 12 minutes it is 5.8 × 10 ²⁰

For 12 seconds it will be 3.48 × 10 ²²

Explanation:

We know that in radioactive disintegration,

Activity = [tex]$\lambda N$[/tex]

Where, [tex]$\lambda =decay constant[/tex]

N is the number of nuclei available for decay, Here we have

[tex]$N=6.02\times {{10}^{23}}$[/tex]  atoms for decay, So

[tex]$\lambda =\frac{\ln 2}{{{t}_{{\scriptstyle{}^{1}/{}_{2}}}}}$[/tex]

(a) Total activity = [tex]$\lambda N$[/tex]  [tex]$=\frac{0.693}{12000\times 31536000}\times 6.023\times {{10}^{23}}decays/\sec $[/tex]

[tex]$=1.1\times {{10}^{12}}$[/tex]

(b) Total activity = [tex]$\lambda N$[/tex]  [tex]$=\frac{0.693}{12\times 3600}\times 6.023\times {{10}^{23}}decays/\sec $[/tex]

[tex]$=9.66\times {{10}^{18}}$[/tex]

(c) Total activity = [tex]$\lambda N$[/tex]  [tex]$=\frac{0.693}{12\times 60}\times 6.023\times {{10}^{23}}decays/\sec $[/tex]

[tex]$=5.8\times {{10}^{20}}$[/tex]

(d) Total activity = [tex]$\lambda N$[/tex]  [tex]$=\frac{0.693}{12}\times 6.023\times {{10}^{23}}decays/\sec $[/tex]

[tex]$=3.48\times {{10}^{22}}$[/tex]

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https://brainly.com/question/10835600

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