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Question

A 1.00106 g1.00 \cdot 10^{-6} \ g sample of nobelium, 102254No_{102}^{254} \mathrm{No}, has a half-life of 55 seconds after it is formed. What is the percentage of 102254No_{102}^{254} \mathrm{No} remaining at the following times?

(a) 5.0 min after it forms

(b) 1.0 h after it forms

Solution

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We have a formula from which we can calculate the ratio of the radioactive substance:

NN0=eλt\dfrac{N}{N_0}=e^{-\lambda {\cdot} t}

where tt is time, λ\lambda is decay constant, N0N_0 is initial concentration and NN is concentration at time tt.

From the half-life we can get λ\lambda:

λ=ln(2)t1/2\lambda= {\dfrac{ln(2)}{t_{1/2}}}

λ=ln(2)55s\lambda= {\dfrac{ln(2)}{55\text{s}}}

λ=0.0126s1\lambda=0.0126\text{s}^{-1}

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