147665
A radioactive element $X$ converts into another stable element $Y$. Half life of $X$ is 2 hrs. Initially only $X$ is present. After time $t$, the ratio of atoms of $X$ and $Y$ is found to be $1: 4$, then $t$ in hours is
1 2
2 4
3 between 4 and 6
4 6
Explanation:
C Let $\mathrm{N}_{0}$ be the number of atoms of $\mathrm{X}$ at time $\mathrm{t}$ $=0$ Then, at $\mathrm{t}=4 \mathrm{hr}$ (two half lives) $\mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{0}}{4} \text { and } \mathrm{N}_{\mathrm{Y}}=\frac{3 \mathrm{~N}_{0}}{4}$ Then, $\quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{3}$ And at time 6 hrs (three half lives) $\mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{0}}{8} \text { and } \mathrm{N}_{\mathrm{Y}}=\frac{7 \mathrm{~N}_{0}}{8}$ Or $\quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{7}$ The given ratio $\frac{1}{4}$ lies between $\frac{1}{3}$ and $\frac{1}{7}$. Therefore, $\mathrm{t}$ lies between $4 \mathrm{hrs}$ and $6 \mathrm{hrs}$
BITSAT-2016
NUCLEAR PHYSICS
147667
If the decay constant of a radioactive substance is $\lambda$, then its half-life and mean life are respectively
1 $\frac{1}{\lambda}$ and $\frac{\log _{\mathrm{e}} 2}{\lambda}$
2 $\frac{\log _{\mathrm{e}} 2}{\lambda}$ and $\frac{1}{\lambda}$
3 $\lambda \log _{\mathrm{e}} 2$ and $\frac{1}{\lambda}$
4 $\frac{\lambda}{\log _{\mathrm{e}} 2}$ and $\frac{1}{\lambda}$
Explanation:
B According to radioactive decay law - $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ We know that after one half-life we get remaining nuclei as half of initial nuclei $\left(\mathrm{N}_{0}\right)$ $\mathrm{N}=\mathrm{N}_{0} / 2$ so using this with $\mathrm{t}=\mathrm{t}_{1 / 2}$ in equation (i), $\frac{\mathrm{N}_{0}}{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda\left(\mathrm{T}_{1 / 2}\right)}$ $\mathrm{e}^{\lambda \mathrm{T}_{1 / 2}}=2$ Taking $\log$ on both side- $\mathrm{T}_{1 / 2}=\frac{\log _{\mathrm{e}} 2}{\lambda}$ Mean life $(\tau)=\frac{1}{\lambda}$ So, Half life and mean life are respectively $\frac{\log _{\mathrm{e}} 2}{\lambda}$ and $\frac{1}{\lambda}$.
UPSEE - 2010
NUCLEAR PHYSICS
147668
Half - life of radioactive substance is $3.20 \mathrm{~h}$. What is the time taken for a $75 \%$ of substance to be used?
1 $6.38 \mathrm{~h}$
2 $12 \mathrm{~h}$
3 $4.18 \mathrm{~h}$
4 $1.2 \mathrm{~h}$
Explanation:
A Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=3.20 \mathrm{~h}$ According to question, percentage of substance left - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=1-\frac{75}{100}=\frac{25}{100}$ $\Rightarrow \quad\frac{1}{4}=\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\therefore \quad\frac{25}{100}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{t}=2 \mathrm{~T}_{1 / 2}$ $=2 \times 3.20$ $=6.40 \mathrm{~h} \approx 6.38 \mathrm{~h}$
UPSEE - 2010
NUCLEAR PHYSICS
147670
In a radioactive material the activity at time $t_{1}$ is $R_{1}$ and at a later time $t_{2}$, it is $R_{2}$. If the decay constant of the material is $\lambda$, then :
A As we know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{R}=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\mathrm{R}_{1}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ $\mathrm{R}_{2}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$ On dividing equation (i) by (ii), we get- $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}}{-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}}$ $\mathrm{R}_{1}=\mathrm{R}_{2} \mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$
UPSEE - 2006
NUCLEAR PHYSICS
147671
A radioactive material has a half - life of $8 \mathrm{yr}$. The activity of the material will decrease to about $1 / 8$ of its original value in :
147665
A radioactive element $X$ converts into another stable element $Y$. Half life of $X$ is 2 hrs. Initially only $X$ is present. After time $t$, the ratio of atoms of $X$ and $Y$ is found to be $1: 4$, then $t$ in hours is
1 2
2 4
3 between 4 and 6
4 6
Explanation:
C Let $\mathrm{N}_{0}$ be the number of atoms of $\mathrm{X}$ at time $\mathrm{t}$ $=0$ Then, at $\mathrm{t}=4 \mathrm{hr}$ (two half lives) $\mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{0}}{4} \text { and } \mathrm{N}_{\mathrm{Y}}=\frac{3 \mathrm{~N}_{0}}{4}$ Then, $\quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{3}$ And at time 6 hrs (three half lives) $\mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{0}}{8} \text { and } \mathrm{N}_{\mathrm{Y}}=\frac{7 \mathrm{~N}_{0}}{8}$ Or $\quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{7}$ The given ratio $\frac{1}{4}$ lies between $\frac{1}{3}$ and $\frac{1}{7}$. Therefore, $\mathrm{t}$ lies between $4 \mathrm{hrs}$ and $6 \mathrm{hrs}$
BITSAT-2016
NUCLEAR PHYSICS
147667
If the decay constant of a radioactive substance is $\lambda$, then its half-life and mean life are respectively
1 $\frac{1}{\lambda}$ and $\frac{\log _{\mathrm{e}} 2}{\lambda}$
2 $\frac{\log _{\mathrm{e}} 2}{\lambda}$ and $\frac{1}{\lambda}$
3 $\lambda \log _{\mathrm{e}} 2$ and $\frac{1}{\lambda}$
4 $\frac{\lambda}{\log _{\mathrm{e}} 2}$ and $\frac{1}{\lambda}$
Explanation:
B According to radioactive decay law - $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ We know that after one half-life we get remaining nuclei as half of initial nuclei $\left(\mathrm{N}_{0}\right)$ $\mathrm{N}=\mathrm{N}_{0} / 2$ so using this with $\mathrm{t}=\mathrm{t}_{1 / 2}$ in equation (i), $\frac{\mathrm{N}_{0}}{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda\left(\mathrm{T}_{1 / 2}\right)}$ $\mathrm{e}^{\lambda \mathrm{T}_{1 / 2}}=2$ Taking $\log$ on both side- $\mathrm{T}_{1 / 2}=\frac{\log _{\mathrm{e}} 2}{\lambda}$ Mean life $(\tau)=\frac{1}{\lambda}$ So, Half life and mean life are respectively $\frac{\log _{\mathrm{e}} 2}{\lambda}$ and $\frac{1}{\lambda}$.
UPSEE - 2010
NUCLEAR PHYSICS
147668
Half - life of radioactive substance is $3.20 \mathrm{~h}$. What is the time taken for a $75 \%$ of substance to be used?
1 $6.38 \mathrm{~h}$
2 $12 \mathrm{~h}$
3 $4.18 \mathrm{~h}$
4 $1.2 \mathrm{~h}$
Explanation:
A Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=3.20 \mathrm{~h}$ According to question, percentage of substance left - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=1-\frac{75}{100}=\frac{25}{100}$ $\Rightarrow \quad\frac{1}{4}=\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\therefore \quad\frac{25}{100}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{t}=2 \mathrm{~T}_{1 / 2}$ $=2 \times 3.20$ $=6.40 \mathrm{~h} \approx 6.38 \mathrm{~h}$
UPSEE - 2010
NUCLEAR PHYSICS
147670
In a radioactive material the activity at time $t_{1}$ is $R_{1}$ and at a later time $t_{2}$, it is $R_{2}$. If the decay constant of the material is $\lambda$, then :
A As we know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{R}=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\mathrm{R}_{1}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ $\mathrm{R}_{2}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$ On dividing equation (i) by (ii), we get- $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}}{-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}}$ $\mathrm{R}_{1}=\mathrm{R}_{2} \mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$
UPSEE - 2006
NUCLEAR PHYSICS
147671
A radioactive material has a half - life of $8 \mathrm{yr}$. The activity of the material will decrease to about $1 / 8$ of its original value in :
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
NUCLEAR PHYSICS
147665
A radioactive element $X$ converts into another stable element $Y$. Half life of $X$ is 2 hrs. Initially only $X$ is present. After time $t$, the ratio of atoms of $X$ and $Y$ is found to be $1: 4$, then $t$ in hours is
1 2
2 4
3 between 4 and 6
4 6
Explanation:
C Let $\mathrm{N}_{0}$ be the number of atoms of $\mathrm{X}$ at time $\mathrm{t}$ $=0$ Then, at $\mathrm{t}=4 \mathrm{hr}$ (two half lives) $\mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{0}}{4} \text { and } \mathrm{N}_{\mathrm{Y}}=\frac{3 \mathrm{~N}_{0}}{4}$ Then, $\quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{3}$ And at time 6 hrs (three half lives) $\mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{0}}{8} \text { and } \mathrm{N}_{\mathrm{Y}}=\frac{7 \mathrm{~N}_{0}}{8}$ Or $\quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{7}$ The given ratio $\frac{1}{4}$ lies between $\frac{1}{3}$ and $\frac{1}{7}$. Therefore, $\mathrm{t}$ lies between $4 \mathrm{hrs}$ and $6 \mathrm{hrs}$
BITSAT-2016
NUCLEAR PHYSICS
147667
If the decay constant of a radioactive substance is $\lambda$, then its half-life and mean life are respectively
1 $\frac{1}{\lambda}$ and $\frac{\log _{\mathrm{e}} 2}{\lambda}$
2 $\frac{\log _{\mathrm{e}} 2}{\lambda}$ and $\frac{1}{\lambda}$
3 $\lambda \log _{\mathrm{e}} 2$ and $\frac{1}{\lambda}$
4 $\frac{\lambda}{\log _{\mathrm{e}} 2}$ and $\frac{1}{\lambda}$
Explanation:
B According to radioactive decay law - $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ We know that after one half-life we get remaining nuclei as half of initial nuclei $\left(\mathrm{N}_{0}\right)$ $\mathrm{N}=\mathrm{N}_{0} / 2$ so using this with $\mathrm{t}=\mathrm{t}_{1 / 2}$ in equation (i), $\frac{\mathrm{N}_{0}}{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda\left(\mathrm{T}_{1 / 2}\right)}$ $\mathrm{e}^{\lambda \mathrm{T}_{1 / 2}}=2$ Taking $\log$ on both side- $\mathrm{T}_{1 / 2}=\frac{\log _{\mathrm{e}} 2}{\lambda}$ Mean life $(\tau)=\frac{1}{\lambda}$ So, Half life and mean life are respectively $\frac{\log _{\mathrm{e}} 2}{\lambda}$ and $\frac{1}{\lambda}$.
UPSEE - 2010
NUCLEAR PHYSICS
147668
Half - life of radioactive substance is $3.20 \mathrm{~h}$. What is the time taken for a $75 \%$ of substance to be used?
1 $6.38 \mathrm{~h}$
2 $12 \mathrm{~h}$
3 $4.18 \mathrm{~h}$
4 $1.2 \mathrm{~h}$
Explanation:
A Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=3.20 \mathrm{~h}$ According to question, percentage of substance left - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=1-\frac{75}{100}=\frac{25}{100}$ $\Rightarrow \quad\frac{1}{4}=\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\therefore \quad\frac{25}{100}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{t}=2 \mathrm{~T}_{1 / 2}$ $=2 \times 3.20$ $=6.40 \mathrm{~h} \approx 6.38 \mathrm{~h}$
UPSEE - 2010
NUCLEAR PHYSICS
147670
In a radioactive material the activity at time $t_{1}$ is $R_{1}$ and at a later time $t_{2}$, it is $R_{2}$. If the decay constant of the material is $\lambda$, then :
A As we know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{R}=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\mathrm{R}_{1}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ $\mathrm{R}_{2}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$ On dividing equation (i) by (ii), we get- $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}}{-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}}$ $\mathrm{R}_{1}=\mathrm{R}_{2} \mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$
UPSEE - 2006
NUCLEAR PHYSICS
147671
A radioactive material has a half - life of $8 \mathrm{yr}$. The activity of the material will decrease to about $1 / 8$ of its original value in :
147665
A radioactive element $X$ converts into another stable element $Y$. Half life of $X$ is 2 hrs. Initially only $X$ is present. After time $t$, the ratio of atoms of $X$ and $Y$ is found to be $1: 4$, then $t$ in hours is
1 2
2 4
3 between 4 and 6
4 6
Explanation:
C Let $\mathrm{N}_{0}$ be the number of atoms of $\mathrm{X}$ at time $\mathrm{t}$ $=0$ Then, at $\mathrm{t}=4 \mathrm{hr}$ (two half lives) $\mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{0}}{4} \text { and } \mathrm{N}_{\mathrm{Y}}=\frac{3 \mathrm{~N}_{0}}{4}$ Then, $\quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{3}$ And at time 6 hrs (three half lives) $\mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{0}}{8} \text { and } \mathrm{N}_{\mathrm{Y}}=\frac{7 \mathrm{~N}_{0}}{8}$ Or $\quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{7}$ The given ratio $\frac{1}{4}$ lies between $\frac{1}{3}$ and $\frac{1}{7}$. Therefore, $\mathrm{t}$ lies between $4 \mathrm{hrs}$ and $6 \mathrm{hrs}$
BITSAT-2016
NUCLEAR PHYSICS
147667
If the decay constant of a radioactive substance is $\lambda$, then its half-life and mean life are respectively
1 $\frac{1}{\lambda}$ and $\frac{\log _{\mathrm{e}} 2}{\lambda}$
2 $\frac{\log _{\mathrm{e}} 2}{\lambda}$ and $\frac{1}{\lambda}$
3 $\lambda \log _{\mathrm{e}} 2$ and $\frac{1}{\lambda}$
4 $\frac{\lambda}{\log _{\mathrm{e}} 2}$ and $\frac{1}{\lambda}$
Explanation:
B According to radioactive decay law - $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ We know that after one half-life we get remaining nuclei as half of initial nuclei $\left(\mathrm{N}_{0}\right)$ $\mathrm{N}=\mathrm{N}_{0} / 2$ so using this with $\mathrm{t}=\mathrm{t}_{1 / 2}$ in equation (i), $\frac{\mathrm{N}_{0}}{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda\left(\mathrm{T}_{1 / 2}\right)}$ $\mathrm{e}^{\lambda \mathrm{T}_{1 / 2}}=2$ Taking $\log$ on both side- $\mathrm{T}_{1 / 2}=\frac{\log _{\mathrm{e}} 2}{\lambda}$ Mean life $(\tau)=\frac{1}{\lambda}$ So, Half life and mean life are respectively $\frac{\log _{\mathrm{e}} 2}{\lambda}$ and $\frac{1}{\lambda}$.
UPSEE - 2010
NUCLEAR PHYSICS
147668
Half - life of radioactive substance is $3.20 \mathrm{~h}$. What is the time taken for a $75 \%$ of substance to be used?
1 $6.38 \mathrm{~h}$
2 $12 \mathrm{~h}$
3 $4.18 \mathrm{~h}$
4 $1.2 \mathrm{~h}$
Explanation:
A Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=3.20 \mathrm{~h}$ According to question, percentage of substance left - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=1-\frac{75}{100}=\frac{25}{100}$ $\Rightarrow \quad\frac{1}{4}=\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\therefore \quad\frac{25}{100}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{t}=2 \mathrm{~T}_{1 / 2}$ $=2 \times 3.20$ $=6.40 \mathrm{~h} \approx 6.38 \mathrm{~h}$
UPSEE - 2010
NUCLEAR PHYSICS
147670
In a radioactive material the activity at time $t_{1}$ is $R_{1}$ and at a later time $t_{2}$, it is $R_{2}$. If the decay constant of the material is $\lambda$, then :
A As we know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{R}=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\mathrm{R}_{1}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ $\mathrm{R}_{2}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$ On dividing equation (i) by (ii), we get- $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}}{-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}}$ $\mathrm{R}_{1}=\mathrm{R}_{2} \mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$
UPSEE - 2006
NUCLEAR PHYSICS
147671
A radioactive material has a half - life of $8 \mathrm{yr}$. The activity of the material will decrease to about $1 / 8$ of its original value in :
147665
A radioactive element $X$ converts into another stable element $Y$. Half life of $X$ is 2 hrs. Initially only $X$ is present. After time $t$, the ratio of atoms of $X$ and $Y$ is found to be $1: 4$, then $t$ in hours is
1 2
2 4
3 between 4 and 6
4 6
Explanation:
C Let $\mathrm{N}_{0}$ be the number of atoms of $\mathrm{X}$ at time $\mathrm{t}$ $=0$ Then, at $\mathrm{t}=4 \mathrm{hr}$ (two half lives) $\mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{0}}{4} \text { and } \mathrm{N}_{\mathrm{Y}}=\frac{3 \mathrm{~N}_{0}}{4}$ Then, $\quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{3}$ And at time 6 hrs (three half lives) $\mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{0}}{8} \text { and } \mathrm{N}_{\mathrm{Y}}=\frac{7 \mathrm{~N}_{0}}{8}$ Or $\quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{7}$ The given ratio $\frac{1}{4}$ lies between $\frac{1}{3}$ and $\frac{1}{7}$. Therefore, $\mathrm{t}$ lies between $4 \mathrm{hrs}$ and $6 \mathrm{hrs}$
BITSAT-2016
NUCLEAR PHYSICS
147667
If the decay constant of a radioactive substance is $\lambda$, then its half-life and mean life are respectively
1 $\frac{1}{\lambda}$ and $\frac{\log _{\mathrm{e}} 2}{\lambda}$
2 $\frac{\log _{\mathrm{e}} 2}{\lambda}$ and $\frac{1}{\lambda}$
3 $\lambda \log _{\mathrm{e}} 2$ and $\frac{1}{\lambda}$
4 $\frac{\lambda}{\log _{\mathrm{e}} 2}$ and $\frac{1}{\lambda}$
Explanation:
B According to radioactive decay law - $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ We know that after one half-life we get remaining nuclei as half of initial nuclei $\left(\mathrm{N}_{0}\right)$ $\mathrm{N}=\mathrm{N}_{0} / 2$ so using this with $\mathrm{t}=\mathrm{t}_{1 / 2}$ in equation (i), $\frac{\mathrm{N}_{0}}{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda\left(\mathrm{T}_{1 / 2}\right)}$ $\mathrm{e}^{\lambda \mathrm{T}_{1 / 2}}=2$ Taking $\log$ on both side- $\mathrm{T}_{1 / 2}=\frac{\log _{\mathrm{e}} 2}{\lambda}$ Mean life $(\tau)=\frac{1}{\lambda}$ So, Half life and mean life are respectively $\frac{\log _{\mathrm{e}} 2}{\lambda}$ and $\frac{1}{\lambda}$.
UPSEE - 2010
NUCLEAR PHYSICS
147668
Half - life of radioactive substance is $3.20 \mathrm{~h}$. What is the time taken for a $75 \%$ of substance to be used?
1 $6.38 \mathrm{~h}$
2 $12 \mathrm{~h}$
3 $4.18 \mathrm{~h}$
4 $1.2 \mathrm{~h}$
Explanation:
A Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=3.20 \mathrm{~h}$ According to question, percentage of substance left - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=1-\frac{75}{100}=\frac{25}{100}$ $\Rightarrow \quad\frac{1}{4}=\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\therefore \quad\frac{25}{100}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{t}=2 \mathrm{~T}_{1 / 2}$ $=2 \times 3.20$ $=6.40 \mathrm{~h} \approx 6.38 \mathrm{~h}$
UPSEE - 2010
NUCLEAR PHYSICS
147670
In a radioactive material the activity at time $t_{1}$ is $R_{1}$ and at a later time $t_{2}$, it is $R_{2}$. If the decay constant of the material is $\lambda$, then :
A As we know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{R}=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\mathrm{R}_{1}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ $\mathrm{R}_{2}=-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$ On dividing equation (i) by (ii), we get- $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}}{-\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}}$ $\mathrm{R}_{1}=\mathrm{R}_{2} \mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$
UPSEE - 2006
NUCLEAR PHYSICS
147671
A radioactive material has a half - life of $8 \mathrm{yr}$. The activity of the material will decrease to about $1 / 8$ of its original value in :