Question

\[90\% \]of a radioactive sample is left undecayed after time \['to\] has elapsed. What percentage of the initial sample will decay in a total time\[2t\]?
A. \[20\% \]
B. \[19\% \]
C. \[40\% \]
D. \[38\% \]

Answer 1

Hint: Though the exact percentage of a radioactive sample that will remain undecayed after a certain amount of time has elapsed can be difficult to predict, it is generally agreed that a large majority of the sample will still be present.

Complete step by step solution:
It is important to note that the term "radioactive" refers to the instability of certain atoms, which causes them to emit energy in the form of radiation. This radiation can be either alpha particles, beta particles, or gamma rays.

The emission of these particles is random, meaning it is impossible to predict when a particular atom will decay. The period it takes for half of the atoms in a sample of a radioactive element to decay is known as the half-life. This means that if the half-life of a particular component is known, it is possible to predict how long it will take for half of the atoms in a sample to decay.

However, it is essential to remember that the half-life is only an average and that the actual Amount of time it takes for any given atom to decay is entirely random. This means that a significant portion of a sample can decay very quickly or, for a minimal amount of a model, decay very slowly.

Given that \[90\% \] is left un-decayed after time \['t'\] .
Hence, \[10\% \]decays in time\[\].
Initial estimates of the substance's quantity are \['x'\]
After time \[\] \[10\% \] is decayed i.e. Amount of substance left \[ = 0.9\]\[x\]
After further time \['t'\]another \[10\% \]is decayed.
i.e. \[0.1 \times 0.9x\]is decayed
Leaving behind\[0.81\]\[x\].
Hence after time \[2t\] we see that \[0.9\]\[x\] has decayed, which is\[19\% \].
Hence, the correct answer is option (B).
Note: The half-life of a radioactive element is only an average, and that the actual Amount of time it takes for any given atom to decay is completely random.