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Question: Ac features a half-life of 22.0 years with reference to radioactive decay. The decay follows two par...

Ac features a half-life of 22.0 years with reference to radioactive decay. The decay follows two parallel paths, one resulting to Th222T{h^{222}} and therefore the other to Fr223F{r^{223}} . The percentage yields to those two daughter nuclides are 2.0 and 98.0 respectively. What are the decay constants ( λ\lambda ) for every of the separate paths?

Explanation

Solution

We must remember that the radioactive element tends to emit radioactive energy in the surrounding. The law of radioactivity gives the relation between the initial amount of element, decay constant and time taken to decay. By using that law of radioactivity, the time taken by the radioactive element to scale back to 1/e times is calculated.

Complete answer:
We have to remember that the radioactive decay is said to be half-life. The time required for an isotope to be reduced to half of its original mass through radioactive decay is the half-life of that substance. Through radioactive decay, unstable isotopes undergo decay by emitting radiation.
Given data contains,
t12=22years{t_{\dfrac{1}{2}}} = 22years
By using half-life, we will calculate the decay constant,
Half-life is denoted as t12t\dfrac{1}{2}.
We can calculate the decay constant by using the given formula as,
λ=0.693t12\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}
On substituting the values we get,
⇒λ=0.69322\Rightarrow \lambda = \dfrac{{0.693}}{{22}}
On simplifying we get,
⇒λ=0.315years\Rightarrow \lambda = 0.315years
Fractional yield of Ac227→A{c^{227}} \to Th nuclide
=2.02.0+98= \dfrac{{2.0}}{{2.0 + 98}}
⇒\RightarrowFractional yield=20100 = \dfrac{{20}}{{100}}
On simplifying we get,
⇒\RightarrowFractional yield=0.02 = 0.02
Now we can calculate the decay constant as,
λ′{\lambda'}= fractional yield \times $$$$\lambda
Substitute the known values in the above equation we get,
⇒λ′=0.02×0.0315\Rightarrow {\lambda'} = 0.02 \times 0.0315
On simplifying we get,
λ′=6.3×10−4year−1\lambda ' = 6.3 \times {10^{ - 4}}yea{r^{ - 1}}
Therefore, the solution to this present question is 6.3×10−4year−16.3 \times {10^{ - 4}}yea{r^{ - 1}} .

Note:
We will find this without doing calculations as we did. We will reduce one half-life of the initial quantity whenever we pass a half-life time. Initially 8 years it becomes half, the subsequent 8years, that is the 16 years afterwards, it becomes one-quarter and this continues up to 32 years when it finally becomes one-sixteenth.