Question

A radioactive element A decays by the sequence and with half-lives given below: \(\mathrm{A} \frac{\alpha}{30 \mathrm{~min}}>\mathrm{B} \frac{2 \beta}{2 \mathrm{days}} \rightarrow \mathrm{C}\) Which one of the following statement is correct? (a) after two hours, less than \(10 \%\) of the initial \(\mathrm{A}\) is left (b) maximum amount of B present at any time is less than \(50 \%\) of the initial amount of \(A\). (c) stomic number of \(\mathrm{A}\) and \(\mathrm{C}\) are same (d) both (a) and (c) are correct

Short answer

Correct statements are (a) and (c), so the answer is (d).

Step-by-step solution

Understand the Decay Sequence

Element A decays into B by emitting an alpha particle and B decays into C by emitting two beta particles. For our purpose, A decreases to B and then B converts to C.

Calculate Half-life Reductions for A in Two Hours

A has a half-life of 30 minutes. To find how much A remains after 2 hours, determine how many half-lives occur in 2 hours: \[ \text{Number of half-lives in 2 hours} = \frac{120 \text{ minutes}}{30 \text{ minutes}} = 4 \]Applying the formula for half-life decay, the amount of A remaining is \( \left( \frac{1}{2} \right)^4 \approx 0.0625 \), or 6.25%.

Determine Element B's Maximum Amount

B starts forming as soon as A starts decaying. Since B has a longer half-life of 2 days, more B is initially formed. However, the maximum amount B can reasonably approach is initially half of the remaining A because B also decays to C. Maximum amount of B is eventually less than 50% before it decays to C.

Analyze Atomic Numbers

A decays to B by emitting an alpha particle (0 ightarrow00) which reduces the atomic number by 2. B then decays to C by emitting beta particles; each emission increases the atomic number by 1. The atomic number changes by: \[ Z_A \rightarrow (Z_A-2) \rightarrow (Z_A-2+2) \]Therefore, the atomic number of C is the same as A.

Conclusion from Steps

Step 2 confirms (a) since less than 10% of A remains after 2 hours and Step 4 confirms (c) since the atomic numbers of A and C are the same.

Key concepts

Half-life Calculation

Half-life is a critical concept in understanding how radioactive materials decay over time. It represents the time it takes for half of a radioactive substance to decay into its products. In our exercise, element A has a half-life of 30 minutes. This means that every 30 minutes, the quantity of A is reduced by half.

To calculate how much of element A remains after a specific period, you can use the formula \[N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}\]where \(N(t)\) is the remaining quantity of A at time \(t\), \(N_0\) is the initial quantity, and \(T_{1/2}\) is the half-life. To apply this in the exercise, we determine how many half-lives occur in two hours (120 minutes): \[\text{Number of half-lives} = \frac{120 \text{ minutes}}{30 \text{ minutes}} = 4\]After four half-lives, the amount of A left is:\[N(t) = N_0 \left( \frac{1}{2} \right)^4 \approx 0.0625 \times N_0\]This results in roughly 6.25% of the original substance remaining.

Decay Sequence

In radioactive decay, elements transform into different elements or isotopes through emission of particles. This is called a decay sequence. For instance, in our exercise, element A undergoes a decay sequence to become element B, and subsequently, B decays into C.

Initially, element A emits an alpha particle to form B. This process of alpha decay results in a decrease in the atomic number by 2. When B forms, it doesn’t remain stable for long and further decays into C by emitting two beta particles. Beta decay results in an increase in the atomic number by 1 for each particle emitted. This chain reaction ensures that radioactive substances continue to transform until a stable isotope is reached.

• Alpha decay: Reduces the atomic number.
• Beta decay: Increases the atomic number.

Understanding these sequences helps make sense of how materials change and eventually stabilize over time.

Atomic Number Changes

Radioactive decay affects an element’s atomic number, determined by the number of protons in its nucleus. This is crucial in understanding the identity and properties of an element. In the decay sequence provided in the exercise, we see how these changes occur.

When element A decays by emitting an alpha particle:

• alpha decay reduces A's atomic number by 2, transitioning A \((Z_A)\) to B \((Z_A - 2)\).

Later, as B undergoes beta decay twice, its atomic number is:

• first incremented by 1, \((Z_A - 2 + 1)\)
• then incremented by another 1, \((Z_A - 2 + 2)\)

This results in the atomic number of element C being the same as the initial element A. Thus, despite the transformations due to emission of particles, the overall series of decays brings us back to the original atomic number, showcasing a balance in nuclear transformation.