The decay constant of radium is 4.28 times 10 | vedclass

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(D) The relationship between the half-life $(T_{1/2})$ and the decay constant $(\lambda)$ is given by the formula: $T_{1/2} = \frac{\ln(2)}{\lambda} \

Vedclass · Accepted Answer

$1620$

Vedclass · Answer

(D) The relationship between the half-life $(T_{1/2})$ and the decay constant $(\lambda)$ is given by the formula: $T_{1/2} = \frac{\ln(2)}{\lambda} \approx \frac{0.6931}{\lambda}$.Given the decay constant $\lambda = 4.28 	imes 10^{-4} 	ext{ year}^{-1}$.Substituting the value into the formula:$T_{1/2} = \frac{0.6931}{4.28 	imes 10^{-4}} 	ext{ years}$.$T_{1/2} = \frac{0.6931}{4.28} 	imes 10^{4} 	ext{ years}$.$T_{1/2} \approx 0.161939 	imes 10^{4} 	ext{ years}$.$T_{1/2} \approx 1619.39 	ext{ years}$.Rounding to the nearest whole number,we get $1620 	ext{ years}$.

The decay constant of radium is $4.28 \times 10^{-4}$ per year. Its half-life will be .......... $years$.

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