Question

8g of $C{u^{66}}$ undergoes radioactive decay and after $15$ minutes only $1g$ remains. The half-life, in minutes, is then?
A) $\dfrac{{15\ln (2)}}{{\ln (8)}}$
B) $\dfrac{{15\ln (8)}}{{\ln (2)}}$
C) $\dfrac{{15}}{8}$
D) $\dfrac{{15}}{8}$
E) $15\ln (2)$

Answer 1

Radioactive decay is purely a nuclear phenomenon and depends on any physical and chemical condition. And the half-life of a radioactive substance is a characteristic constant. It measures the time it takes for a given amount of the substance to become reduced by half.

Complete step by step solution:
The radioactive decay law is a universal law that describes the statistical behaviour of a large number of nuclides. And the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the given sample material.
According to the question,
Initial mass of radioactive $C{u^{66}}$ sample ${m_o} = 8g$
Mass of the substance remains $m = 1g$
Time of decay of $C{u^{66}} = 15\min$
Radioactive decay follows first order kinetic, the rate of decay is proportional to the number of un-decay atom in radioactive substance at any time$t$,
First order kinetic rate law,
$\dfrac{{dN}}{{dt}} = - \lambda N$
${N_t} = {N_o}{e^{ - \lambda t}}$
Now,
$\lambda t = \ln \dfrac{{{N_o}}}{{{N_t}}} = \dfrac{m}{{{m_o}}}$
We have given that the time is $15\min$ for decay,
Substitute the value of time and initial and final amount of the radioactive substance.
$\lambda (15) = \ln \dfrac{1}{8}$
$\lambda = \dfrac{{\ln \left( 8 \right)}}{{15}}$ ……………(1)
And calculate the half- life of the radioactive sample or substance,
${t_{\dfrac{1}{2}}} = \dfrac{{\ln (2)}}{\lambda }$ ……………….(2)
Now, substitute the value of half-life from equation (1) into equation (2),
We get,
${t_{\dfrac{1}{2}}} = \dfrac{{\ln (2) \times 15}}{{\ln (8)}}$

Hence, The correct option of this question is (A).

Note: Decay constant of radioactive sample is equal to reciprocal of the time after which the number of remaining active atoms reduces to $\dfrac{1}{e}$ time of original value.