The half-life of a radioactive substance agai | vedclass

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(D) The decay rate is given by the formula: $\frac{dN}{dt} = \lambda N$.First, calculate the decay constant $\lambda$ using the half-life $T_{1/2} = 1

Vedclass · Accepted Answer

$2.3 	imes 10^8 \, 	ext{atoms/s}$

Vedclass · Answer

(D) The decay rate is given by the formula: $\frac{dN}{dt} = \lambda N$.First, calculate the decay constant $\lambda$ using the half-life $T_{1/2} = 1.2 	imes 10^7 \, s$:$\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{1.2 	imes 10^7} \, s^{-1}$.Now, substitute the values of $\lambda$ and $N = 4.0 	imes 10^{15}$ atoms into the decay rate formula:$\frac{dN}{dt} = \left( \frac{0.693}{1.2 	imes 10^7} ight) 	imes (4.0 	imes 10^{15})$.$\frac{dN}{dt} = \frac{0.693 	imes 4.0}{1.2} 	imes 10^{15-7}$.$\frac{dN}{dt} = \frac{2.772}{1.2} 	imes 10^8$.$\frac{dN}{dt} = 2.31 	imes 10^8 \, 	ext{atoms/s}$.Rounding to the nearest given option, the decay rate is $2.3 	imes 10^8 \, 	ext{atoms/s}$.

The half-life of a radioactive substance against $\alpha$-decay is $1.2 \times 10^7 \, s$. What is the decay rate for $4.0 \times 10^{15}$ atoms of the substance?

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