A radioactive isotope X has a half-life of 3 seconds. At t = 0 second, a given sample of this isotope X contains 8000 atoms. Calculate (a) its decay constant, (b) the time t_1'when 1000 atoms of the isotope X remain in the sample and (c) the number of decays per second in the sample at t = t₁ second.

Solution

Half life of isotope X = 3 seconds.
Number of atoms at t=0 is 8000 atoms.

(a) Decay constant,
   $λ = \frac{0.6931}{T} = \frac{0.6931}{3} = 0.231 s^{- 1}$

(b) Atoms remaining in the sample = 1000

Time t₁ at which these many sample is remaining = ?

Now, using the formula of the law of radioactive decay we have,

   $t_{1} = \frac{1}{λ} \log_{e} \frac{N_{0}}{N} = \frac{2.3026}{0.231} \log_{10} 8 = 9 s$

(c) Number of decays per second in the sample is,
   $- \frac{dN}{dt} = λN = 0.231 \times 1000 = 231 s^{- 1}$

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