Question

after one hour, 1/16 of the initial amount of certain radioactive isotope remains undecayed. the half life of the isotope is:

Answer 1

Answer:

$t_{1/2} = 0.25 hours$

Explanation:

As we know that

$N = N_o e^{-\lambda t}$

here we know that

$\lambda = \frac{ln2}{t_{1/2}}$

so we will have

$N = N_o e^{\frac{t ln2}{t_{1/2}}}$

so it will be

$N = N_o(\frac{1}{2})^{\frac{t}{t_{1/2}}}$

now we know that

$t = 1 hour$

$N = \frac{N_o}{16}$

$\frac{1}{16} = (\frac{1}{2})^{\frac{t}{t_{1/2}}}$

$4 = \frac{1}{t_{1/2}}$

$t_{1/2} = 0.25 hours$

Answer 2

(1/2)^x=1/16
ln 1/2^x=x ln 1/2=ln 1/16
x=4
4 half-lives in one hour=1/4 hour for the half-life of the isotope
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