
A 16 gm sample of a radioactive element was taken from Bombay to Delhi in 2 hour and it was found that 1 gm of the element remained (disintegrated). Half-life of the element is?
A. 2 hours
B. 1 hours
C. $\dfrac{1}{2}$ hours
D. $\dfrac{1}{4}$ hours
Answer
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Hint:The half-life of a radioactive sample is the amount of time needed for half of its atomic nuclei to spontaneously transform into other nuclear species and release particles and energy, or, more precisely, the amount of time needed for its disintegrations per second to fall by half. In this problem, we have to find the time taken for the sample to decay half the amount i.e, the half life.
Formula used:
We have the equation:
$\dfrac{N}{N_{0}}=\left(\dfrac{1}{2}\right)^{\dfrac{t}{T}}$
Where, $\mathrm{N}$ is the amount of sample remaining after $\mathrm{t}$ time, $N_{0}$ is the original amount of sample, $\mathrm{T}$ is the half-life of the element and $\mathrm{t}$ is the time taken to decay.
Complete step by step solution:
Here 16 gm of a radioactive element is taken from one place to another. After reaching the destination after 2 hrs only 1 gm of the element is remaining. From this data, we have to find the half-life of the element.
Given that, Undecayed sample or remaining amount of sample, $N=1 \mathrm{gm}$
Original amount of sample present, $N_{0}=16 g \mathrm{~m}$
Time taken for the sample to reach from $16 \mathrm{gm}$ to $1 \mathrm{gm}, t=2 \mathrm{hrs}$
We have to find the half-life of the sample.
For that we use the equation,
$\dfrac{N}{N_{0}}=\left(\dfrac{1}{2}\right)^{\dfrac{t}{T}}$
On substituting the given values in the question,
We get:
$\dfrac{1}{16}=\left(\dfrac{1}{2}\right)^{\dfrac{2}{T}}$
$T$ is the half-life of the element that we have to find.
On solving equation by taking natural logarithm on both sides we get:
$\ln \dfrac{1}{16}=\dfrac{2}{T} \ln \dfrac{1}{2}$
From this we get half-life of the element as:
$\therefore T=\dfrac{2}{4}=\dfrac{1}{2} hrs$
Therefore, the answer is option C.
Notes: Here the amount of undecayed or remaining sample is directly given. Therefore, we can use the equation directly. Also, while taking logarithm remember that we have to take natural logarithm.
Formula used:
We have the equation:
$\dfrac{N}{N_{0}}=\left(\dfrac{1}{2}\right)^{\dfrac{t}{T}}$
Where, $\mathrm{N}$ is the amount of sample remaining after $\mathrm{t}$ time, $N_{0}$ is the original amount of sample, $\mathrm{T}$ is the half-life of the element and $\mathrm{t}$ is the time taken to decay.
Complete step by step solution:
Here 16 gm of a radioactive element is taken from one place to another. After reaching the destination after 2 hrs only 1 gm of the element is remaining. From this data, we have to find the half-life of the element.
Given that, Undecayed sample or remaining amount of sample, $N=1 \mathrm{gm}$
Original amount of sample present, $N_{0}=16 g \mathrm{~m}$
Time taken for the sample to reach from $16 \mathrm{gm}$ to $1 \mathrm{gm}, t=2 \mathrm{hrs}$
We have to find the half-life of the sample.
For that we use the equation,
$\dfrac{N}{N_{0}}=\left(\dfrac{1}{2}\right)^{\dfrac{t}{T}}$
On substituting the given values in the question,
We get:
$\dfrac{1}{16}=\left(\dfrac{1}{2}\right)^{\dfrac{2}{T}}$
$T$ is the half-life of the element that we have to find.
On solving equation by taking natural logarithm on both sides we get:
$\ln \dfrac{1}{16}=\dfrac{2}{T} \ln \dfrac{1}{2}$
From this we get half-life of the element as:
$\therefore T=\dfrac{2}{4}=\dfrac{1}{2} hrs$
Therefore, the answer is option C.
Notes: Here the amount of undecayed or remaining sample is directly given. Therefore, we can use the equation directly. Also, while taking logarithm remember that we have to take natural logarithm.
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