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Question: The half-life of a radioactive element is 8 years. How much amount will be present after 32 years? ...

The half-life of a radioactive element is 8 years. How much amount will be present after 32 years?

& \text{A) }\dfrac{1}{4} \\\ & \text{B) }\dfrac{1}{8} \\\ & \text{C) }\dfrac{1}{16} \\\ & \text{D) }\dfrac{1}{32} \\\ \end{aligned}$$
Explanation

Solution

We can very easily find the substance left over after years if we know the half-life time of the radioactive element. The key idea is that after each half-life time half of the initial sample gets converted or gets decayed. We can just do the sum even without the formulae.

Complete answer:
Radioactive elements are those substances which decay into other elements due to its inability to continue in its form or due to its atomic instability. These elements have a specific time period in which half of the total sample gets decayed. This is called the half-life time. This is derived from the radioactive decay law as –
R=R0eλtR={{R}_{0}}{{e}^{-\lambda t}}
Where R is the rate of radioactivity,
R0{{R}_{0}} is the initial rate of radioactivity,
λ\lambda is the disintegration constant or decay constant,
t is the time taken
Now, we can write this in terms of the number of radioactive samples as –
N=N0eλtN={{N}_{0}}{{e}^{-\lambda t}}
Where, N is the number of radioactive elements left over after ‘t’ time,
N0{{N}_{0}}is the initial number of radioactive elements.
The disintegration constant is the reciprocal of the mean time period of decay of particles.
Also, the half-life time of the element is related to the disintegration constant as –

& {{T}_{1/2}}=\dfrac{\ln 2}{\lambda }=\dfrac{0.693}{\lambda } \\\ & \Rightarrow \text{ }\lambda =\dfrac{0.693}{{{T}_{1/2}}} \\\ \end{aligned}$$ Now, let us substitute this in the decay law – $$\begin{aligned} & N={{N}_{0}}{{e}^{-\lambda t}} \\\ & \Rightarrow \text{ }N={{N}_{0}}{{e}^{-\dfrac{0.693}{{{T}_{1/2}}}t}} \\\ & \text{here, } \\\ & {{T}_{1/2}}=8\text{years} \\\ & t=32\text{years} \\\ & \Rightarrow \text{ }N={{N}_{0}}{{e}^{-\dfrac{0.693}{8}\times 32}} \\\ & \Rightarrow \text{ }N={{N}_{0}}{{e}^{-2.772}} \\\ & \Rightarrow \text{ }N=\dfrac{{{N}_{0}}}{16} \\\ \end{aligned}$$ From the above calculations, we understand that the remaining number of radioactive elements is one-sixteenth of the initial number. **So, the correct answer is “Option C”.** **Note:** We can find this without doing the calculations as we did here. We can reduce one-half of the initial quantity each time we pass a half-life time. At first 8 years it becomes half, the next 8 years, i.e., the 16 years afterwards, it becomes one-quarter and this continues up to 32 years when it finally becomes one-sixteenth.