The half-life of radioactive substance $^{238}{92}U$ undergoing $\alpha$-decay is $4.5 imes 10^9$ years. Calculate: The decay constant $\lambda$ in $s^{-1}$. The activity of 1 gram of $^{238}U$ sample. (Given: Avogadro number $NA = 6.023 imes 10^{23}$ atoms/mol, 1 year = $3.15 imes 10^7$ s)

Q: The half-life of radioactive substance $^{238}{92}U$ undergoing $\alpha$-decay is $4.5 \times 10^9$ years. Calculate: The decay constant $\lambda$ in $s^{-1}$. The activity of 1 gram of $^{238}U$ sample. (Given: Avogadro number $NA = 6.023 \times 10^{23}$ atoms/mol, 1 year = $3.15 \times 10^7$ s)

Given Data: Half-life ($T{1/2}$) = $4.5 \times 10^9$ years 1 year = $3.15 \times 10^7$ seconds Mass of Uranium sample ($m$) = 1 g Molar mass of $^{238}U$ ($M$) = 238 g/mol Avogadro number ($NA$) = $6.023 \times 10^{23}$ atoms/mol --- Step 1: Calculation of Decay Constant ($\lambda$ in $s^{-1}$)\nThe relationship between half-life ($T{1/2}$) and decay constant ($\lambda$) is given by: $$\lambda = \frac{\ln(2)}{T{1/2}}$$ \nFirst, convert the half-life from years to seconds: $$T{1/2} = 4.5 \times 10^9 \text{ years} \times 3.15 \times 10^7 \text{ s/year}$$ $$T{1/2} = 1.4175 \times 10^{17} \text{ seconds}$$ \nSubstitute the value of $\ln(2) \approx 0.693$: $$\lambda = \frac{0.693}{1.4175 \times 10^{17} \text{ s}}$$ $$\lambda \approx 4.889 \times 10^{-18} \text{ s}^{-1}$| --- Step 2: Calculation of Number of Nuclei ($N$) in 1 gram of $^{238}U$\nThe number of atoms in 1 gram of the sample is calculated using the formula: $$N = \frac{m \times NA}{M}$| $$N = \frac{1 \text{ g} \times 6.023 \times 10^{23} \text{ atoms/mol}}{238 \text{ g/mol}}$$ $$N = 2.5307 \times 10^{21} \text{ atoms}$$ --- Step 3: Calculation of Activity ($R$)\nActivity represents the rate of decay, given by the radioactive decay law: $$R = \lambda N$$ \nSubstitute the values of $\lambda$ and $N$ calculated above: $$R = (4.899 \times 10^{-18} \text{ s}^{-1}) \times (2.5307 \times 10^{21})$$ $$R = 1.238 \times 10^{4} \text{ disintegrations/second (Bq)}$$ \nTo express this in Curies (Ci) where $1 \text{ Ci} = 3.7 \times 10^{10} \text{ Bq}$: $$R = \frac{1.238 \times 10^{4}}{3.7 \times 10^{10}} \text{ Ci} \approx 3.35 \times 10^{-7} \text{ Ci}$$ Final Answer: The decay constant $\lambda = 4.89 \times 10^{-18} \text{ s}^{-1}$ The activity of 1 gram of $^{238}U$ is $1.24 \times 10^{4} \text{ Bq}$ (or $3.35 \times 10^{-7} \text{ Ci}$)

Question

The half-life of radioactive substance $^{238}{92}U$ undergoing $\alpha$-decay is $4.5 	imes 10^9$ years. Calculate: The decay constant $\lambda$ in $s^{-1}$. The activity of 1 gram of $^{238}U$ sample. (Given: Avogadro number $NA = 6.023 	imes 10^{23}$ atoms/mol, 1 year = $3.15 	imes 10^7$ s)

Accepted Answer

Given Data: Half-life ($T{1/2}$) = $4.5 	imes 10^9$ years 1 year = $3.15 	imes 10^7$ seconds Mass of Uranium sample ($m$) = 1 g Molar mass of $^{238}U$ ($M$) = 238 g/mol Avogadro number ($NA$) = $6.023 	imes 10^{23}$ atoms/mol --- Step 1: Calculation of Decay Constant ($\lambda$ in $s^{-1}$)
The relationship between half-life ($T{1/2}$) and decay constant ($\lambda$) is given by: $$\lambda = \frac{\ln(2)}{T{1/2}}$$ 
First, convert the half-life from years to seconds: $$T{1/2} = 4.5 	imes 10^9 	ext{ years} 	imes 3.15 	imes 10^7 	ext{ s/year}$$ $$T{1/2} = 1.4175 	imes 10^{17} 	ext{ seconds}$$ 
Substitute the value of $\ln(2) \approx 0.693$: $$\lambda = \frac{0.693}{1.4175 	imes 10^{17} 	ext{ s}}$$ $$\lambda \approx 4.889 	imes 10^{-18} 	ext{ s}^{-1}$| --- Step 2: Calculation of Number of Nuclei ($N$) in 1 gram of $^{238}U$
The number of atoms in 1 gram of the sample is calculated using the formula: $$N = \frac{m 	imes NA}{M}$| $$N = \frac{1 	ext{ g} 	imes 6.023 	imes 10^{23} 	ext{ atoms/mol}}{238 	ext{ g/mol}}$$ $$N = 2.5307 	imes 10^{21} 	ext{ atoms}$$ --- Step 3: Calculation of Activity ($R$)
Activity represents the rate of decay, given by the radioactive decay law: $$R = \lambda N$$ 
Substitute the values of $\lambda$ and $N$ calculated above: $$R = (4.899 	imes 10^{-18} 	ext{ s}^{-1}) 	imes (2.5307 	imes 10^{21})$$ $$R = 1.238 	imes 10^{4} 	ext{ disintegrations/second (Bq)}$$ 
To express this in Curies (Ci) where $1 	ext{ Ci} = 3.7 	imes 10^{10} 	ext{ Bq}$: $$R = \frac{1.238 	imes 10^{4}}{3.7 	imes 10^{10}} 	ext{ Ci} \approx 3.35 	imes 10^{-7} 	ext{ Ci}$$ Final Answer: The decay constant $\lambda = 4.89 	imes 10^{-18} 	ext{ s}^{-1}$ The activity of 1 gram of $^{238}U$ is $1.24 	imes 10^{4} 	ext{ Bq}$ (or $3.35 	imes 10^{-7} 	ext{ Ci}$)

Given Data:

Step 1: Calculation of Decay Constant ($\lambda$ in $s^{-1}$)\nThe relationship between half-life ($T_{1/2}$) and decay constant ($\lambda$) is given by:

Step 2: Calculation of Number of Nuclei ($N$) in 1 gram of $^{238}U$\nThe number of atoms in 1 gram of the sample is calculated using the formula:

Step 3: Calculation of Activity ($R$)\nActivity represents the rate of decay, given by the radioactive decay law:

Final Answer: