(C) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{3 \text{ days}}{1 \text{ day}} = 3$. The a

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(C) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{3 \text{ days}}{1 \text{ day}} = 3$. The amount remaining $(N)$ is given by the formula $N = N_o \times (1/2)^n$. Substituting $n = 3$,we get $N = N_o \times (1/2)^3 = N_o \times 1/8$. Therefore,$1/8$ of the original amount remains.

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

Medium

$A$ radioactive element has a half-life of $1 \text{ day}$. After $3 \text{ days}$,the amount of the element left will be:

A
$1/2$ of the original amount
B
$1/4$ of the original amount
C
$1/8$ of the original amount
D
$1/16$ of the original amount

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Nuclear Chemistry

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Rate of decay and Half-life

$A$ sample is obtained from sea water. $11.9$ $C^{14}$ atoms dissociate from $1 \ g$ of carbon. $15.3$ $C^{14}$ atoms dissociate from a live sample. How many years old is the sample? $(t_{1/2} = 5730 \ \text{years})$

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After $2 \, \text{hours}$,the amount of a certain radioactive substance reduces to $1/16^{th}$ of the original amount (the decay process follows first-order kinetics). The half-life of the radioactive substance is $...... \, \text{min}$.

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For a radioactive substance with a half-life period of $500$ years,the time required for the complete decay of $100 \ mg$ of it would be:

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If the disintegration constant is $6.93 \times 10^{-6} \ s^{-1}$,then the half-life of $_{6}C^{14}$ will be:

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$A$ radioactive substance has a half-life $(t_{1/2})$ of $60 \ minutes$. After $3 \ hours$,what percentage of the radioactive substance will remain?

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$A$ radioactive element has a half-life of $1 \text{ day}$. After $3 \text{ days}$,the amount of the element left will be:

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$A$ sample is obtained from sea water. $11.9$ $C^{14}$ atoms dissociate from $1 \ g$ of carbon. $15.3$ $C^{14}$ atoms dissociate from a live sample. How many years old is the sample? $(t_{1/2} = 5730 \ \text{years})$

After $2 \, \text{hours}$,the amount of a certain radioactive substance reduces to $1/16^{th}$ of the original amount (the decay process follows first-order kinetics). The half-life of the radioactive substance is $...... \, \text{min}$.

For a radioactive substance with a half-life period of $500$ years,the time required for the complete decay of $100 \ mg$ of it would be:

If the disintegration constant is $6.93 \times 10^{-6} \ s^{-1}$,then the half-life of $_{6}C^{14}$ will be:

$A$ radioactive substance has a half-life $(t_{1/2})$ of $60 \ minutes$. After $3 \ hours$,what percentage of the radioactive substance will remain?

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