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Two radioactive nuclei A and B are present in equal numbers to begin with. Three days later, there are 2.44 times as many A nuclei as there are B nuclei. The half-life of species B is 1.14 days. Find the half-life of species A (in days).

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Answer:

2.23 days

Explanation:

Given that;

N=Noe^-kt

Where;

N= number of radioactive nuclei present at time =t

No= number of radioactive nuclei originally present

k = decay constant

t = time taken

But k = ln2/T

Where T is the half life of the radioactive nuclei

Hence;

NA= Noe^-ln2t/TA     (1)

NB= Noe^-ln2t/TB      (2)

But NA =2.44NB

2.44NB = Noe^-ln2t/TA     (3)

NB= Noe^-ln2t/TB      (4)

Divide  (3) by  (4)

2.44NB/NB = Noe^-ln2t/TA/Noe^-ln2t/TB

2.44 = e^-ln2t/TA + ln2t/TB

Taking natural logarithm of both sides

ln2.44 = -ln2t/TA + ln2t/TB

Given that TB = 1.14 days

TA= -ln2t(ln2.44 -ln2t/TB )^-1

TA= -(2.079) (0.892 - 2.079/1.14)^-1

TA = 2.23 days

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