Two radioactive samples have decay constant 15x and 3x. If they have same number of nuclei initially, the ratio of number of nuclei after a time $\dfrac {1}{6x}$ is
Answer
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Hint: To solve this problem, use the formula for number of nuclei at time t. Using the formula, find the number of nuclei at time $\dfrac {1}{6x}$ for both the radioactive samples. Then, take the ratio of the number of nuclei at time $\dfrac {1}{6x}$ for the first sample to the number of nuclei at time $\dfrac {1}{6x}$ for the second sample. This will give the ratio of the number of nuclei after a time $\dfrac {1}{6x}$.
Formula used:
$N= {N}_{0}{e}^{- \lambda t}$
Complete answer:
Given: Decay constant of first sample (${\lambda}_{1}$)= 15x
Decay constant of second sample (${\lambda}_{2}$)= 3x
The number of nuclei at time t is given by,
$N= {N}_{0}{e}^{- \lambda t}$
Where, ${N}_{0}$ is the number of nuclei at t=0
t is the time
$\lambda$ is the decay constant
The number of nuclei at time $\dfrac {1}{6x}$ for first sample is given by,
${N}_{1}= {N}_{0}{e}^{- {\lambda}_{1} t}$
Substituting values in above equation we get,
${N}_{1}= {N}_{0}{e}^{-\dfrac {15x}{16}}$
$\Rightarrow {N}_{1}= {N}_{0}{e}^{-2.5x}$ ...(1)
Similarly, the number of nuclei at time $\dfrac {1}{6x}$ for second sample is given by,
${N}_{2}= {N}_{0}{e}^{- {\lambda}_{2} t}$
Substituting values in above equation we get,
${N}_{2}= {N}_{0}{e}^{-\dfrac {3x}{16}}$
$\Rightarrow {N}_{2}= {N}_{0}{e}^{-0.5x}$N ...(2)
Now, taking the ratio of equations. (1) and (2) we get,
$N= \dfrac {{N}_{1}}{{N}_{2}}$
Substituting values in above equation we get,
$N= \dfrac {{N}_{0}{e}^{-2.5x}}{{N}_{0}{e}^{-0.5x}}$
$\Rightarrow N= \dfrac {{e}^{-2.5x}}{{e}^{-0.4x}}$
$\Rightarrow N= {e}^{-2.5} \times {e}^{0.5}$
$\Rightarrow N= {e}^{-2}$
$\therefore N= \dfrac {1}{{e}^{2}}$
Hence, the ratio of the number of nuclei after a time $\dfrac {1}{6x}$ is $ \dfrac {1}{{e}^{2}}$.
Note:
In order to solve such types of problems, students must remember the basic definition and these relations. Decay constant is the reciprocal of time during which the number of nuclei of a radioactive substance decreases to $\dfrac {1}{e}$ or 36.8% of the initial number of nuclei. The decay constant varies for different types of nuclei. Decay constant depends upon various environmental factors such as temperature, pressure etc.
Formula used:
$N= {N}_{0}{e}^{- \lambda t}$
Complete answer:
Given: Decay constant of first sample (${\lambda}_{1}$)= 15x
Decay constant of second sample (${\lambda}_{2}$)= 3x
The number of nuclei at time t is given by,
$N= {N}_{0}{e}^{- \lambda t}$
Where, ${N}_{0}$ is the number of nuclei at t=0
t is the time
$\lambda$ is the decay constant
The number of nuclei at time $\dfrac {1}{6x}$ for first sample is given by,
${N}_{1}= {N}_{0}{e}^{- {\lambda}_{1} t}$
Substituting values in above equation we get,
${N}_{1}= {N}_{0}{e}^{-\dfrac {15x}{16}}$
$\Rightarrow {N}_{1}= {N}_{0}{e}^{-2.5x}$ ...(1)
Similarly, the number of nuclei at time $\dfrac {1}{6x}$ for second sample is given by,
${N}_{2}= {N}_{0}{e}^{- {\lambda}_{2} t}$
Substituting values in above equation we get,
${N}_{2}= {N}_{0}{e}^{-\dfrac {3x}{16}}$
$\Rightarrow {N}_{2}= {N}_{0}{e}^{-0.5x}$N ...(2)
Now, taking the ratio of equations. (1) and (2) we get,
$N= \dfrac {{N}_{1}}{{N}_{2}}$
Substituting values in above equation we get,
$N= \dfrac {{N}_{0}{e}^{-2.5x}}{{N}_{0}{e}^{-0.5x}}$
$\Rightarrow N= \dfrac {{e}^{-2.5x}}{{e}^{-0.4x}}$
$\Rightarrow N= {e}^{-2.5} \times {e}^{0.5}$
$\Rightarrow N= {e}^{-2}$
$\therefore N= \dfrac {1}{{e}^{2}}$
Hence, the ratio of the number of nuclei after a time $\dfrac {1}{6x}$ is $ \dfrac {1}{{e}^{2}}$.
Note:
In order to solve such types of problems, students must remember the basic definition and these relations. Decay constant is the reciprocal of time during which the number of nuclei of a radioactive substance decreases to $\dfrac {1}{e}$ or 36.8% of the initial number of nuclei. The decay constant varies for different types of nuclei. Decay constant depends upon various environmental factors such as temperature, pressure etc.
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