A radioactive element has a half-life of 200 | vedclass

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(C) The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{200} \approx 0.003465 \ day^{-1}$.The activity $A$ at tim

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$75$

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(C) The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{200} \approx 0.003465 \ day^{-1}$.The activity $A$ at time $t$ is given by $A = A_0 e^{-\lambda t}$.$\frac{A}{A_0} = e^{-\lambda t} = e^{-(0.693/200) 	imes 83} = e^{-0.2877}$.Alternatively,using the formula $A = A_0 (1/2)^{t/t_{1/2}}$:$\frac{A}{A_0} = (0.5)^{83/200} = (0.5)^{0.415}$.Taking $\log$ on both sides: $\log(\frac{A}{A_0}) = 0.415 	imes \log(0.5) = 0.415 	imes (-0.301) \approx -0.1249$.$\frac{A}{A_0} = 	ext{antilog}(-0.1249) = 10^{-0.1249} \approx 0.75$.Percentage remaining = $0.75 	imes 100 = 75 \ \%$.

$A$ radioactive element has a half-life of $200 \ days$. The percentage of original activity remaining after $83 \ days$ is $....$ (Nearest integer).
(Given: $\text{antilog } 0.125 = 1.333$,$\text{antilog } 0.693 = 4.93$)

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$A$ radioactive element has a half-life of $200 \ days$. The percentage of original activity remaining after $83 \ days$ is $....$ (Nearest integer).(Given: $\text{antilog } 0.125 = 1.333$,$\text{antilog } 0.693 = 4.93$)