Question

Two radioactive isotopes A and B of atomic mass X and Y are mixed in equal amount by mass. After $20$ days, their mass ratio is found to be $1:4$ . Half-life of A is one day. What will be the half-life of B?
A) $1.11$ day.
B) $0.6237$ day.
C) $0.11\dfrac{X}{Y}$ day.
D) $1.11\dfrac{Y}{X}$ day.

Answer 1

We know that the half-life of a radioactive isotope is the time it takes for one-portion of the sample to decay. The half-existence of a radioactive isotope is a property of a given isotope and is independent of the measure of test, temperature, and weight. In this way, if the half-life and measure of sample are known, it is conceivable to anticipate the amount of the radioactive isotope will stay after a timeframe.

Complete step by step answer:
We know that the mass of the isotope is proportional to the activity of the isotope which in turn proportional to the number of nuclei in the isotope. The half-life period of the substance is given the formula,
$\lambda = \dfrac{{2.303}}{t}\log \dfrac{N}{{{N_a}}}$
Where $\lambda$ the decay constant, t is is the time period. N and Na are the amount of substance present initially and the amounts of substance which remain after time period respectively.
Let us assume one gram of A and B are taken at first. ${W_A}$ and ${W_B}$ are the amounts of substance which remain after twenty days.
${\lambda _A} = \dfrac{{2.303}}{{20}}\log \dfrac{1}{{{W_A}}}$
${\lambda _B} = \dfrac{{2.303}}{{20}}\log \dfrac{1}{{{W_B}}}$
${\lambda _A} - {\lambda _B} = \dfrac{{2.303}}{{20}}\log \dfrac{{{W_B}}}{{{W_A}}}$
Let’s we substitute the given values we get,
$\Rightarrow {\lambda _A} - {\lambda _B} = \dfrac{{2.303}}{{20}}\log \dfrac{4}{1} = 0.0693$
${\lambda _B} = {\lambda _A} - 0.0693$
$\Rightarrow {\lambda _B} = \dfrac{{0.693}}{{{t_{1/2}}}} - 0.0693 = \dfrac{{0.693}}{1} - 0.0693$
On simplification we get,
$\Rightarrow {t_{1/2}}B = \dfrac{{0.0693}}{{0.6237}} = 1.11$
The half-life of B is $1.11$ day.

So, the correct answer is Option A.

Note: Now we discuss about the radioactive law as,
We have to remember that the radioactive law expresses that the probability per unit time that a nucleus will decompose is a steady free of time.
$N = {N_0}.{e^{ - kt}}$
Where N is the number of radioactive nuclei left unreacted time t. N0 is the number of radioactive nuclei present in the beginning. The decay constant is given as k. Time required to decay is t.
Let’s we can see the Half-life period as,
We must know that the half-life is characterized as the measure of time it takes for an offered isotope to lose half of its radioactivity.
${t_{\left( {1/2} \right)}} = \dfrac{{\ln 2}}{k}$
Where k is the decay constant.