Respuesta :
notice q=3/2, is half of the originalĀ q = 3(1/2)t/28.8
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Explanation:
Mass of the strontium-90 present at initial stage= [tex]N_o=3 grams[/tex]
Mass of the strontium-90 present after time t = N = q
The half life of strontium-90 = [tex]t_{1/2}=28.8 years[/tex]
Decay constant = [tex]\lambda [/tex]
[tex]\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{28.8 years}=0.02406 years^{-1}[/tex]
[tex]\ln N=\ln N_o -\lambda t[/tex]
[tex]\ln q=\ln 3 - 0.02406 years^{-1}\times t[/tex]
[tex]0.02406 years^{-1}\times t=\ln 3 - \ln q[/tex]
[tex]0.02406 years^{-1}\times t=\ln \frac{3}{q}[/tex]
[tex]t=41.50 \ln \frac{3}{q}[/tex]
[tex] e^t=(\frac{3}{q})^{41.50}[/tex]
From the above expression we can see that t and q are inversely related to each other.