(A) The number of half-lives $(n)$ is calculated as: $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{18 \ h}{3 \ h} = 6$.Using the radioactive

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(A) The number of half-lives $(n)$ is calculated as: $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{18 \ h}{3 \ h} = 6$.Using the radioactive decay formula: $N_t = N_o \times (1/2)^n$.Substituting the values: $N_t = 256 \times (1/2)^6$.$N_t = 256 \times \frac{1}{64} = 4 \ g$.Therefore,the remaining mass is $4.0 \ g$.

Class 12 Chemistry Nuclear Chemistry Rate of decay and Half-life

Medium

The half-life of a radioactive isotope is $3 \ hours$. If the initial mass of the isotope were $256 \ g$,the mass of it remaining undecayed after $18 \ hours$ would be ....... $g$. (in $.0$)

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A
$4$
B
$8$
C
$12$
D
$16$

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Nuclear Chemistry

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Rate of decay and Half-life

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The half-life of a radioactive isotope is $3 \ hours$. If the initial mass of the isotope were $256 \ g$,the mass of it remaining undecayed after $18 \ hours$ would be ....... $g$. (in $.0$)

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If $n_t$ number of radioatoms are present at time $t$, the following expression will be a constant:

$A$ radioactive isotope has a $t_{1/2}$ of $10 \ days$. If today $125 \ g$ of it is left,what was its weight $40 \ days$ earlier?

The half-life of radioactive $C^{14}$ is $5760 \ years$. In how many years will a $200 \ mg$ sample of $C^{14}$ be reduced to $25 \ mg$?

The half-life of $1 \ g$ of a radioactive sample is $9 \ hours$. The radioactive decay obeys first-order kinetics. The time required for the original sample to reduce to $0.2 \ g$ is .......... $hours$.

$A$ radioactive sample $(Z = 22)$ decreases by $90\%$ in $10 \ \text{years}$. What will be the half-life of the sample in $\text{years}$?

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