The half- life of a certain radioactive element is $1$ year. The $3$ percentage of the radioactive element is decayed in $6$ months will be
A. $29.3\% $
B. $6739.5\% $
C. $70.5\% $
D. $14.5\% $
Answer
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Hint: The number of atoms in the radioactive sample in terms of the initial number of atoms have to be found. For this the decay constant can be determined from the half-life of the sample. The decay constant is inversely proportional to the half-life of the sample.
Complete step-by-step solution:
The expression for number of atoms present in the radioactive element after time $t$ is given as,
${N_t} = {N_0}{e^{ - \lambda t}}..........\left( 1 \right)$
Where, ${N_0}$is the number of atoms present at time $t = 0$, $\lambda $ is the decay constant .
And the expression connecting the half-life and decay constant is given as,
$\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$
Where, ${t_{\dfrac{1}{2}}}$is the half -life of the radioactive sample.
The half -life of the radioactive sample is the time taken for the radioactive sample to degenerate to half of its actual number of atoms.
Given the half- life is $1$ year. Therefore the decay constant is
$\lambda = \dfrac{{0.693}}{1} = 0.693$
From equation $\left( 1 \right)$
$\dfrac{{{N_t}}}{{{N_0}}} = {e^{ - \lambda t}}$
Substitute the values for $6$ months that is $\dfrac{1}{2}{\text{year}}$ .
$
\dfrac{{{N_t}}}{{{N_0}}} = {e^{ - 0.693 \times \dfrac{1}{2}}} = 0.707 \\
{N_t} = 0.707{N_0} \\
$
The equation for calculating the percentage decayed is given as,
$
\% \;{\text{decayed}} = \dfrac{{{N_0} - {N_t}}}{{{N_0}}} \times 100 \\
= \dfrac{{{N_0} - 0.707{N_0}}}{{{N_0}}} \times 100 \\
= \dfrac{{0.293{N_0}}}{{{N_0}}} \times 100 \\
= 29.3\% \\
$
Thus the percentage of the radioactive element decayed in $6$ month is $29.3\% $.
The answer is option A.
Note:- We have to note that the decay constant will be the same for an element for any time. As the time increases exponentially the number of atoms remaining will decrease drastically. And the percentage of decay is calculated with respect to the initial amount of sample.
Complete step-by-step solution:
The expression for number of atoms present in the radioactive element after time $t$ is given as,
${N_t} = {N_0}{e^{ - \lambda t}}..........\left( 1 \right)$
Where, ${N_0}$is the number of atoms present at time $t = 0$, $\lambda $ is the decay constant .
And the expression connecting the half-life and decay constant is given as,
$\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$
Where, ${t_{\dfrac{1}{2}}}$is the half -life of the radioactive sample.
The half -life of the radioactive sample is the time taken for the radioactive sample to degenerate to half of its actual number of atoms.
Given the half- life is $1$ year. Therefore the decay constant is
$\lambda = \dfrac{{0.693}}{1} = 0.693$
From equation $\left( 1 \right)$
$\dfrac{{{N_t}}}{{{N_0}}} = {e^{ - \lambda t}}$
Substitute the values for $6$ months that is $\dfrac{1}{2}{\text{year}}$ .
$
\dfrac{{{N_t}}}{{{N_0}}} = {e^{ - 0.693 \times \dfrac{1}{2}}} = 0.707 \\
{N_t} = 0.707{N_0} \\
$
The equation for calculating the percentage decayed is given as,
$
\% \;{\text{decayed}} = \dfrac{{{N_0} - {N_t}}}{{{N_0}}} \times 100 \\
= \dfrac{{{N_0} - 0.707{N_0}}}{{{N_0}}} \times 100 \\
= \dfrac{{0.293{N_0}}}{{{N_0}}} \times 100 \\
= 29.3\% \\
$
Thus the percentage of the radioactive element decayed in $6$ month is $29.3\% $.
The answer is option A.
Note:- We have to note that the decay constant will be the same for an element for any time. As the time increases exponentially the number of atoms remaining will decrease drastically. And the percentage of decay is calculated with respect to the initial amount of sample.
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