Radioactive material 'A' has decay constant '8λ' and material 'B' has decay constant 'λ'. Initially they have the same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A' will be e?
A. $\dfrac{1}{{9\lambda }} $
B. $\dfrac{1}{\lambda } $
C. $\dfrac{1}{{7\lambda }}$
D. $\dfrac{1}{{8\lambda }} $
Answer
586.5k+ views
Hint:We are given with two radioactive materials with their decay constants. Here, we will use the formula of the number of particles which are present in a decay. The two radioactive materials will provide two different equations and from these two equations, we will divide them to obtain the time taken to get a ratio of e.
Complete step by step answer:
Radioactivity is a process through which certain naturally occurring or artificial nuclides can undergo spontaneous decay releasing new energy. So, this decay process has been accompanied by the emission of one or more types of radiation ionizing or non-ionized particles. Radioactive materials can unstable nuclei that emit ionizing radiation as it decays. As we know, the number of nuclei having decay constant and life time is given by the following formula:
\[N = {N_0}{e^{( - \lambda t)}}\]
Here, N is the final number of atoms, ${N_0}$ is the number of initial atoms, λ is the decay constant and t is the lifetime.
Here, we have two materials with decay constants 8 λ and λ for materials A and B respectively. Let initially they have the same number of nuclei ${N_0}$. Let ${N_A}$ and ${N_B}$ be the final number of atoms for the materials A and B. Thus, we have
\[{N_A} = {N_0}{e^{( - 8\lambda t)}} \\
\Rightarrow{N_B} = {N_0}{e^{( - \lambda t)}} \\
\]
Thus, we are given in the question that;
\[\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{1}{e}\\
\Rightarrow {e^{( - 7\lambda t)}} = {e^{( - 1)}} \\
\Rightarrow 7\lambda t = 1 \\
\therefore t = \dfrac{1}{{7\lambda }} \\
\]
Thus, option C is the correct answer.
Note:There are three types of radioactive decay which being the alpha decay, beta decay and the gamma decay. Radioactive materials are quite dangerous. The mean lifetime of radioactive material is quite small. When a radioactive particle interacts with another atom, the interaction causes the other atom to lose electrons hence becomes ionized.
Complete step by step answer:
Radioactivity is a process through which certain naturally occurring or artificial nuclides can undergo spontaneous decay releasing new energy. So, this decay process has been accompanied by the emission of one or more types of radiation ionizing or non-ionized particles. Radioactive materials can unstable nuclei that emit ionizing radiation as it decays. As we know, the number of nuclei having decay constant and life time is given by the following formula:
\[N = {N_0}{e^{( - \lambda t)}}\]
Here, N is the final number of atoms, ${N_0}$ is the number of initial atoms, λ is the decay constant and t is the lifetime.
Here, we have two materials with decay constants 8 λ and λ for materials A and B respectively. Let initially they have the same number of nuclei ${N_0}$. Let ${N_A}$ and ${N_B}$ be the final number of atoms for the materials A and B. Thus, we have
\[{N_A} = {N_0}{e^{( - 8\lambda t)}} \\
\Rightarrow{N_B} = {N_0}{e^{( - \lambda t)}} \\
\]
Thus, we are given in the question that;
\[\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{1}{e}\\
\Rightarrow {e^{( - 7\lambda t)}} = {e^{( - 1)}} \\
\Rightarrow 7\lambda t = 1 \\
\therefore t = \dfrac{1}{{7\lambda }} \\
\]
Thus, option C is the correct answer.
Note:There are three types of radioactive decay which being the alpha decay, beta decay and the gamma decay. Radioactive materials are quite dangerous. The mean lifetime of radioactive material is quite small. When a radioactive particle interacts with another atom, the interaction causes the other atom to lose electrons hence becomes ionized.
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