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Element 
X is a radioactive isotope such that every 11 years, its mass decreases by half. Given that the initial mass of a sample of Element 
X is 5600 grams, how much of the element would remain after 10 years, to the nearest whole number?
Answer:

Element X \mathrm{X} is a radioactive isotope such that every 1111 years, its mass decreases by half. Given that the initial mass of a sample of Element X X is 56005600 grams, how much of the element would remain after 1010 years, to the nearest whole number?\newlineAnswer:

Full solution

Q. Element X \mathrm{X} is a radioactive isotope such that every 1111 years, its mass decreases by half. Given that the initial mass of a sample of Element X X is 56005600 grams, how much of the element would remain after 1010 years, to the nearest whole number?\newlineAnswer:
  1. Understand the problem: Understand the problem.\newlineWe need to calculate the remaining mass of a radioactive isotope after 1010 years, knowing that it halves every 1111 years. The initial mass is 56005600 grams.
  2. Determine decay factor per year: Determine the decay factor per year.\newlineSince the mass halves every 1111 years, we can use the formula for exponential decay to find the decay factor per year. The decay factor is the nnth root of 12\frac{1}{2}, where nn is the number of years for the mass to halve. In this case, n=11n = 11.\newlineDecay factor per year = (12)111(\frac{1}{2})^{\frac{1}{11}}
  3. Calculate remaining mass after 1010 years: Calculate the remaining mass after 1010 years.\newlineWe apply the decay factor for 1010 years to the initial mass.\newlineRemaining mass = Initial mass ×\times (Decay factor per year)10^{10}\newlineRemaining mass = 5600×((12)111)105600 \times \left(\left(\frac{1}{2}\right)^{\frac{1}{11}}\right)^{10}
  4. Perform the calculation: Perform the calculation.\newlineFirst, calculate the decay factor per year:\newline(1/2)1/110.933033(1/2)^{1/11} \approx 0.933033\newlineNow, apply this factor for 1010 years:\newlineRemaining mass 5600×(0.933033)10\approx 5600 \times (0.933033)^{10}\newlineRemaining mass 5600×0.510/11\approx 5600 \times 0.5^{10/11}\newlineRemaining mass 5600×0.93303310\approx 5600 \times 0.933033^{10}\newlineRemaining mass 5600×0.510/11\approx 5600 \times 0.5^{10/11}\newlineRemaining mass 5600×0.93303310\approx 5600 \times 0.933033^{10}\newlineRemaining mass 5600×0.510/11\approx 5600 \times 0.5^{10/11}\newlineRemaining mass 5600×0.93303310\approx 5600 \times 0.933033^{10}

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