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A F^32 radio nuclide with half-life T=14...

A `F^32` radio nuclide with half-life `T=14.3` days is produced in a reactor at a constant rate `q=2xx10^9` nuclei per second. How soon after the beginning of production of that radio nuclide will its activity be equal to `R=10^9` disintegration per second?

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
`N=R/lambda=((10^9)/(0.693))/(14.3xx3600)=7.43xx10^13`
Now, `(dN)/(dt)=q-lambdaN` or `int_0^N(dN)/(q-lambdaN)=int_0^tdt`
`:.` `N=q/lambda(1-e^(-lambdat))`
Substituting the values,
`7.43xx10^13=((2xx10^9)/(0.693))/(14.3xx3600)[1-e^(-(0.693//14.3xx3600)t)]`
Solving this equation we get,
`t=14.3h`
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Knowledge Check

  • A .^(32)P radionuclide with half - life T = 14.3 days is produced in a reactor at a constant rate q=2.7xx10^(9) nuclei per second. How soon after the beginning of production of that nuclide will its activity be equal to A=1.0xx10^(9)dis.//s ?

    A
    9.5 days
    B
    8 days
    C
    7.5 days
    D
    6 days

  • A radionuclide with half - life 1620 s is produced in a reactor at a constant rate of 1000 nuclei per second. During each decay energy, 200 MeV is released. If the production of radionuclides started at t = 0, then the rate of release of energy at t = 3240 s is

    A
    `1.5xx10^(5)" MeV s"^(-1)`
    B
    `1.5xx10^(2)" MeV s"^(-1)`
    C
    `2.5xx10^(2)" MeV s"^(-1)`
    D
    `3.5xx10^(5)" MeV s"^(-1)`

  • If the rate constant of a reaction is 2xx10^(-3) per second. What is the order of a reaction ?

    A
    0
    B
    1
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    2
    D
    3