
A radioactive element emits 200 particles per second. After three hours 25 particles per second are emitted. The half life of the element will be
A) 50 minutes
B) 60 minutes
C) 70 minutes
D) 80 minutes
Answer
164.4k+ views
Hint:The problem is from the radioactivity section of physics. We need to apply the concepts and equation of activity to solve this problem.
Formula used:
$\dfrac{N}{{{N_0}}} = \dfrac{1}{{{2^n}}}$
Where, $N$= Number of nuclei after time t, ${N_0}$= Initial number of nuclei and $n$= number of half-life.
$n = \dfrac{t}{{{T_{1/2}}}}$
Where t = time taken for decay and ${T_{1/2}}$= Half-life of the radioactive element
Complete answer:
Activity (R) is the number of disintegrations per second of the radioactive sample. Becquerel (Bq) is the unit of activity.
$R = \dfrac{{dN}}{{dt}} \Rightarrow R \propto N$
${R_1} = 200$ particles per second
${R_2} = 25$ particles per second
$\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{{N_2}}}{{{N_1}}}$
${N_1}$ = number of samples when activity is ${R_1}$
${N_2}$ = number of samples when activity is ${R_2}$
We know the relation: $\dfrac{{{N_2}}}{{{N_1}}} = \dfrac{1}{{{2^n}}}$. Where n is the number of half-lives and calculated by $n = \dfrac{t}{{{T_{1/2}}}}$. Where t = time and ${T_{1/2}}$= half-life.
$\dfrac{{{N_2}}}{{{N_1}}} = \dfrac{1}{{{2^n}}} \Rightarrow \dfrac{{{R_2}}}{{{R_1}}} = \dfrac{1}{{{2^n}}}$
$\dfrac{{25}}{{200}} = \dfrac{1}{{{2^n}}}$
$\dfrac{1}{8} = \dfrac{1}{{{2^n}}}$
$\dfrac{1}{{{2^3}}} = \dfrac{1}{{{2^n}}} \Rightarrow n = 3$
Apply this in the equation of n and find half-life. Time is given, t = 3 hours.
$n = \dfrac{t}{{{T_{1/2}}}} \Rightarrow {T_{1/2}} = \dfrac{t}{n}$
${T_{1/2}} = \dfrac{{3hr}}{3} = 1hr$
Hence, the correct option is Option (B).
Additional Information:
The law of radioactive decay states, “If a radioactive sample contains N nuclei, at a given instant the ratio of the radioactive decay ($ - \dfrac{{dN}}{{dt}}$) to the number of nuclei present at that instant is a constant.”
A half-life is the amount of time needed for radioactivity to decline and drop to half of its starting value.
Note:A radioactive source's activity, which is measured as the rate of isotope decay, is what determines how powerful it is. It refers specifically to the quantity of atoms that decay and release radiation every second.
Formula used:
$\dfrac{N}{{{N_0}}} = \dfrac{1}{{{2^n}}}$
Where, $N$= Number of nuclei after time t, ${N_0}$= Initial number of nuclei and $n$= number of half-life.
$n = \dfrac{t}{{{T_{1/2}}}}$
Where t = time taken for decay and ${T_{1/2}}$= Half-life of the radioactive element
Complete answer:
Activity (R) is the number of disintegrations per second of the radioactive sample. Becquerel (Bq) is the unit of activity.
$R = \dfrac{{dN}}{{dt}} \Rightarrow R \propto N$
${R_1} = 200$ particles per second
${R_2} = 25$ particles per second
$\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{{N_2}}}{{{N_1}}}$
${N_1}$ = number of samples when activity is ${R_1}$
${N_2}$ = number of samples when activity is ${R_2}$
We know the relation: $\dfrac{{{N_2}}}{{{N_1}}} = \dfrac{1}{{{2^n}}}$. Where n is the number of half-lives and calculated by $n = \dfrac{t}{{{T_{1/2}}}}$. Where t = time and ${T_{1/2}}$= half-life.
$\dfrac{{{N_2}}}{{{N_1}}} = \dfrac{1}{{{2^n}}} \Rightarrow \dfrac{{{R_2}}}{{{R_1}}} = \dfrac{1}{{{2^n}}}$
$\dfrac{{25}}{{200}} = \dfrac{1}{{{2^n}}}$
$\dfrac{1}{8} = \dfrac{1}{{{2^n}}}$
$\dfrac{1}{{{2^3}}} = \dfrac{1}{{{2^n}}} \Rightarrow n = 3$
Apply this in the equation of n and find half-life. Time is given, t = 3 hours.
$n = \dfrac{t}{{{T_{1/2}}}} \Rightarrow {T_{1/2}} = \dfrac{t}{n}$
${T_{1/2}} = \dfrac{{3hr}}{3} = 1hr$
Hence, the correct option is Option (B).
Additional Information:
The law of radioactive decay states, “If a radioactive sample contains N nuclei, at a given instant the ratio of the radioactive decay ($ - \dfrac{{dN}}{{dt}}$) to the number of nuclei present at that instant is a constant.”
A half-life is the amount of time needed for radioactivity to decline and drop to half of its starting value.
Note:A radioactive source's activity, which is measured as the rate of isotope decay, is what determines how powerful it is. It refers specifically to the quantity of atoms that decay and release radiation every second.
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