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Current Question (ID: 18348)

Question:

$\text{During a nuclear explosion, one of the products is } ^{90}\text{Sr with a half-life of 28.1 years.}$ $\text{If 1 } \mu\text{g of } ^{90}\text{Sr was absorbed in the bones of a newly born baby instead of calcium,}$ $\text{the amount of } ^{90}\text{Sr that will remain after 10 years in the now grown up child would be -}$ $\text{(Given, antilog(0.108) = 1.28)}$

Options:

1. $0.227 \ \mu\text{g}$
2. $0.781 \ \mu\text{g}$
3. $7.81 \ \mu\text{g}$
4. $2.27 \ \mu\text{g}$

Solution:

$\text{HINT: Radioactive decay follow first order kinetics.}$ $\text{Explanation:}$ $\text{Step 1: Calculate the value of rate constant } k \text{ as follows:}$ $k = \frac{0.693}{t_{1/2}}$ $= \frac{0.693}{28.1} \ \text{years}^{-1}$ $k = 0.025 \ \text{years}^{-1}$ $\text{Step 2:}$ $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$ $0.025 = \frac{2.303}{10} \log \frac{[R]_0}{[R]}$ $0.108 = \log \frac{1}{[R]}$ $\text{antilog} (0.108) = \frac{1}{[R]}$ $1.28 = \frac{1}{[R]}$ $[R] = \frac{1}{1.28}$ $[R] = 0.78 \ \mu\text{g}$ $\text{Therefore, } 0.781 \ \mu\text{g of } ^{90}\text{Sr will remain after 10 years.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}