Rankers Physics

Uncategorized: Practice Problem & Solution

53. A sample of radioactive element containing $4 \times 10^{16}$ active nuclei. Half life of element is 10 days, then number of decayed nuclei after 30 days: (2002)
$0.5 \times 10^{16}$
$2 \times 10^{16}$
$3.5 \times 10^{16}$
$1 \times 10^{16}$

Solution Explained:

To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:

30 days represents 3 half-lives ($30/10 = 3$). The number of active nuclei remaining is $N = N_0 (1/2)^3 = (4 \times 10^{16})/8 = 0.5 \times 10^{16}$. The number of decayed nuclei is $4 \times 10^{16} - 0.5 \times 10^{16} = 3.5 \times 10^{16}$.

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