
A radioisotope has a half-life of 75 years. The fraction of the atoms of this material that would decay in 150 years will be
A. 66.6%
B. 85.5%
C. 62.5%
D. 75%
Answer
161.1k+ views
Hint:This problem can be solved by using the concept of radioactivity. We have to use the relation between half life and number of half life. Half-life is the time interval during which half of the atoms of the radioactive sample decay.
Formula Used:
The number of radioactive nuclei remaining after n half-life = \[\dfrac{{{N_{_0}}}}{{{2^n}}}\].
Number of half-life, $n = \dfrac{t}{{{T_{1/2}}}}$
Where ${T_{1/2}}$ = half-life, ${N_0}$= Initial number of nuclei, t = time and n = number of half-lives.
Complete step by step solution:
The half-life of radioisotope is given. That is ${T_{1/2}} = 75$ years. Also t = 150 years.
Number of half-life, $n = \dfrac{t}{{{T_{1/2}}}}$
\[n = \dfrac{{150}}{{75}} = 2\]
Fraction of atoms decayed = \[{N_0} - \left( {\dfrac{{{N_0}}}{{{2^2}}}} \right) = \dfrac{{2{N_0} - {N_0}}}{4} = \dfrac{3}{4} = 0.75 \]
In terms of percentage it is 75%.
Hence, the correct option is option D.
Additional Information: The equation for half-life is given below.
${T_{1/2}} = \dfrac{{0.693}}{\lambda }$
Where, ${T_{1/2}}$= Half-Life and $\lambda $= Disintegration constant.
The law of radioactive decay states, “If a radioactive sample contains N nuclei, at a given instant the ratio of the radioactive decay ($ - \dfrac{{dN}}{{dt}}$) to the number of nuclei present at that instant is a constant.”
\[\dfrac{{( - \dfrac{{dN}}{{dt}})}}{N} = \lambda \\\]
\[\Rightarrow \dfrac{{dN}}{{dt}} = - \lambda N\]
The equation can be re-written as,
${N_t} = {N_0}{e^{ - \lambda t}}$
Where, ${N_t}$ = Number of nuclei present after some time t
${N_0}$= Initial number of nuclei present
\[\lambda \]= Disintegration constant
t = Time
Note: An atom that has become stable in terms of energy by emitting radiation will no longer emit radiation. The amount of radioactive nuclei decreases over time and hence the radioactivity weakens. Radioactive decay is a statistical process.
Formula Used:
The number of radioactive nuclei remaining after n half-life = \[\dfrac{{{N_{_0}}}}{{{2^n}}}\].
Number of half-life, $n = \dfrac{t}{{{T_{1/2}}}}$
Where ${T_{1/2}}$ = half-life, ${N_0}$= Initial number of nuclei, t = time and n = number of half-lives.
Complete step by step solution:
The half-life of radioisotope is given. That is ${T_{1/2}} = 75$ years. Also t = 150 years.
Number of half-life, $n = \dfrac{t}{{{T_{1/2}}}}$
\[n = \dfrac{{150}}{{75}} = 2\]
Fraction of atoms decayed = \[{N_0} - \left( {\dfrac{{{N_0}}}{{{2^2}}}} \right) = \dfrac{{2{N_0} - {N_0}}}{4} = \dfrac{3}{4} = 0.75 \]
In terms of percentage it is 75%.
Hence, the correct option is option D.
Additional Information: The equation for half-life is given below.
${T_{1/2}} = \dfrac{{0.693}}{\lambda }$
Where, ${T_{1/2}}$= Half-Life and $\lambda $= Disintegration constant.
The law of radioactive decay states, “If a radioactive sample contains N nuclei, at a given instant the ratio of the radioactive decay ($ - \dfrac{{dN}}{{dt}}$) to the number of nuclei present at that instant is a constant.”
\[\dfrac{{( - \dfrac{{dN}}{{dt}})}}{N} = \lambda \\\]
\[\Rightarrow \dfrac{{dN}}{{dt}} = - \lambda N\]
The equation can be re-written as,
${N_t} = {N_0}{e^{ - \lambda t}}$
Where, ${N_t}$ = Number of nuclei present after some time t
${N_0}$= Initial number of nuclei present
\[\lambda \]= Disintegration constant
t = Time
Note: An atom that has become stable in terms of energy by emitting radiation will no longer emit radiation. The amount of radioactive nuclei decreases over time and hence the radioactivity weakens. Radioactive decay is a statistical process.
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