
The half-life of a radioactive substance is 48 hours. How much time will it take to disintegrate to its \[\frac{1}{16}\,th\] part
A. 12 hours
B. 16 hours
C. 48 hours
D. 192 hours
Answer
164.1k+ views
Hint: Half-life period of a radioactive substance is defined as the time required for one-half of the radioactive substance to disintegrate. Use this to find out the disintegrate constant. To find out the time required for one sixteenth of the substance to decay, we have to find how much substance will be left after one sixteenth of it has disintegrated. Then equate these values using decay law to find out the value for time taken.
Formula used:
$\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}$
Where: \[{N_0} = \] = Initial concentration, \[{N} = \] = Concentration after time t
And T be the half-life of a radioactive substance.
Complete answer:
Radioactive elements are mainly made up of atoms whose nuclei are unstable and give off atomic radiation in a process of attaining stability. The emitting radiation transformed from radioactive atoms into another chemical element. It may be stable or radioactive such that it undergoes further decay.
In an element of a long period, each disintegrating atom emits only one particle. Hence the number of particles counted must be equal to the number of atoms disintegrating.
The time interval required for half of the atomic nuclei of a radioactive substance to disintegration.
Alpha and gamma radiation is the most common radiation. Radioactive disintegration rates normally state in terms of their half-lives. Decay constant is the probability of disintegration per unit time.
$\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}$
Given value is $\dfrac{N}{{{N_0}}}$= \[\frac{1}{16}\,th\]
$N = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}$
$\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}$ here, T = 48 hours
Here, substitute the values we got above:
$\left( {\dfrac{1}{{16}}} \right) = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{48}}}}$
${\left( {\dfrac{1}{2}} \right)^4} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{48}}}}$
Then, equate the powers of both sides we get,
$4 = \dfrac{t}{{48}}$
We need to find out t, so we get:
$t = 4 \times 48$
$t = 192h$
That is one sixteenth of the radioactive sample will decay in 192 hours
Therefore, option D is the correct option.
Note: In the radioactive nucleus the time taken to become half is known as the half-life. To calculate the time for disintegration of radioactive substances, the ratio of the substance remaining to the initial amount of substance is taken. In the first half-life the maximum amount of substance disintegrates.
Formula used:
$\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}$
Where: \[{N_0} = \] = Initial concentration, \[{N} = \] = Concentration after time t
And T be the half-life of a radioactive substance.
Complete answer:
Radioactive elements are mainly made up of atoms whose nuclei are unstable and give off atomic radiation in a process of attaining stability. The emitting radiation transformed from radioactive atoms into another chemical element. It may be stable or radioactive such that it undergoes further decay.
In an element of a long period, each disintegrating atom emits only one particle. Hence the number of particles counted must be equal to the number of atoms disintegrating.
The time interval required for half of the atomic nuclei of a radioactive substance to disintegration.
Alpha and gamma radiation is the most common radiation. Radioactive disintegration rates normally state in terms of their half-lives. Decay constant is the probability of disintegration per unit time.
$\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}$
Given value is $\dfrac{N}{{{N_0}}}$= \[\frac{1}{16}\,th\]
$N = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}$
$\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}$ here, T = 48 hours
Here, substitute the values we got above:
$\left( {\dfrac{1}{{16}}} \right) = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{48}}}}$
${\left( {\dfrac{1}{2}} \right)^4} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{48}}}}$
Then, equate the powers of both sides we get,
$4 = \dfrac{t}{{48}}$
We need to find out t, so we get:
$t = 4 \times 48$
$t = 192h$
That is one sixteenth of the radioactive sample will decay in 192 hours
Therefore, option D is the correct option.
Note: In the radioactive nucleus the time taken to become half is known as the half-life. To calculate the time for disintegration of radioactive substances, the ratio of the substance remaining to the initial amount of substance is taken. In the first half-life the maximum amount of substance disintegrates.
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