147901
The half-life of ${ }^{60} \mathrm{Co}$ is approximately 5.25 years. In a sample containing $1 \mathrm{gm}$ of freshly prepared ${ }^{60} \mathrm{Co}$, how much of the isotope will be left after 21 years?
1 $125 \mathrm{mg}$
2 $62.5 \mathrm{mg}$
3 nothing will be left
4 $31.25 \mathrm{mg}$
Explanation:
B Given that, half-life of ${ }^{60} \mathrm{Co}\left(\mathrm{T}_{1 / 2}\right)=5.25 \mathrm{yr}$ Initial number of sample $\left(\mathrm{N}_{0}\right)=1 \mathrm{~g}$ Total time taken $(\mathrm{t})=21$ year We know that, Number of half-life $(\mathrm{n})=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{21}{5.25}=4$ Amount of sample left - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=1 \times\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}=\frac{1}{16} \mathrm{~g}=0.0625 \mathrm{~g}$ $\mathrm{~N}=62.5 \mathrm{mg}$
J and K CET-2012
NUCLEAR PHYSICS
147902
What amount of original radioactive material is left after 3 half-lives?
1 $6.5 \%$
2 $12.5 \%$
3 $25.5 \%$
4 $33.3 \%$
Explanation:
B Given that, Original amount of the radioactive material $=\mathrm{N}_{0}$ Number of half-life (n) $=3$ Amount of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{8}$ Hence, the percentage of original amount, $\mathrm{N}=\frac{\mathrm{N}_{0}}{8} \times 100$ $\mathrm{~N}=12.5 \% \text { of } \mathrm{N}_{0}$
J and K CET-2015
NUCLEAR PHYSICS
147903
The number of atoms of a radioactive substance of half-life $T$ is $N_{0}$ at $t=0$. The time necessary to decay from $N_{0} / 2$ atoms to $N_{0} / 10$ atoms will be
1 $\frac{5}{2} \mathrm{~T}$
2 $\mathrm{T} \log 5$
3 $\mathrm{T} \log \left[\frac{5}{2}\right]$
4 $\frac{T}{2} \log 5$
Explanation:
C Given that, half-life $(\mathrm{T})=\frac{\ln 2}{\lambda}$ $\lambda=\frac{\ln 2}{\mathrm{~T}}$ We know that, law of radioactivity - $\mathrm{N}=\mathrm{N}_{\mathrm{o}}^{-\lambda \mathrm{t}}$ According to question - $\frac{\mathrm{N}_{\mathrm{o}}}{2}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ $\frac{1}{2}=\mathrm{e}^{-\lambda t_{1}}$ Taking $\log$ both the side, we get - $\lambda t_{1}=\ln 2$ Similarly, $\frac{\mathrm{N}_{\mathrm{o}}}{10}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$ $\frac{1}{10}=\mathrm{e}^{-\lambda \mathrm{t}_{2}}$ $\lambda \mathrm{t}_{2}=\ln 10$ Subtracting equation (ii) from equation (i), we get, $\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)=\ln 10-\ln 2$ $\frac{\ln 2}{\mathrm{~T}}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)=\ln 10-\ln 2$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\frac{\mathrm{T}(\ln 10-\ln 2)}{\ln 2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T}\left(\frac{\ln \left(\frac{10}{2}\right)}{\ln 2}\right)$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T} \frac{\ln 5}{\ln 2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T} \ln \left(\frac{5}{2}\right)$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T} \log \left(\frac{5}{2}\right)$
WB JEE 2013
NUCLEAR PHYSICS
147904
The half-life of radioactive material is $3 \mathrm{~h}$. If the initial amount is $300 \mathrm{~g}$. Then, after $18 \mathrm{~h}$, it will remain.
1 $4.69 \mathrm{~g}$
2 $46.8 \mathrm{~g}$
3 $9.375 \mathrm{~g}$
4 93.75
Explanation:
A Given that, Half-life of radioactive material $\left(\mathrm{T}_{1 / 2}\right)=3 \mathrm{~h}$ Initial amount of radioactive material $\left(\mathrm{N}_{\mathrm{O}}\right)=300 \mathrm{~g}$ Total time taken $(\mathrm{t})=18 \mathrm{~h}$ Amount of radioactive material left, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{18}{3}}$ $\mathrm{~N}=300 \times\left(\frac{1}{2}\right)^{6}$ $\mathrm{~N}=300 \times \frac{1}{64}$ $\mathrm{~N}=4.69 \mathrm{~g}$
147901
The half-life of ${ }^{60} \mathrm{Co}$ is approximately 5.25 years. In a sample containing $1 \mathrm{gm}$ of freshly prepared ${ }^{60} \mathrm{Co}$, how much of the isotope will be left after 21 years?
1 $125 \mathrm{mg}$
2 $62.5 \mathrm{mg}$
3 nothing will be left
4 $31.25 \mathrm{mg}$
Explanation:
B Given that, half-life of ${ }^{60} \mathrm{Co}\left(\mathrm{T}_{1 / 2}\right)=5.25 \mathrm{yr}$ Initial number of sample $\left(\mathrm{N}_{0}\right)=1 \mathrm{~g}$ Total time taken $(\mathrm{t})=21$ year We know that, Number of half-life $(\mathrm{n})=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{21}{5.25}=4$ Amount of sample left - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=1 \times\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}=\frac{1}{16} \mathrm{~g}=0.0625 \mathrm{~g}$ $\mathrm{~N}=62.5 \mathrm{mg}$
J and K CET-2012
NUCLEAR PHYSICS
147902
What amount of original radioactive material is left after 3 half-lives?
1 $6.5 \%$
2 $12.5 \%$
3 $25.5 \%$
4 $33.3 \%$
Explanation:
B Given that, Original amount of the radioactive material $=\mathrm{N}_{0}$ Number of half-life (n) $=3$ Amount of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{8}$ Hence, the percentage of original amount, $\mathrm{N}=\frac{\mathrm{N}_{0}}{8} \times 100$ $\mathrm{~N}=12.5 \% \text { of } \mathrm{N}_{0}$
J and K CET-2015
NUCLEAR PHYSICS
147903
The number of atoms of a radioactive substance of half-life $T$ is $N_{0}$ at $t=0$. The time necessary to decay from $N_{0} / 2$ atoms to $N_{0} / 10$ atoms will be
1 $\frac{5}{2} \mathrm{~T}$
2 $\mathrm{T} \log 5$
3 $\mathrm{T} \log \left[\frac{5}{2}\right]$
4 $\frac{T}{2} \log 5$
Explanation:
C Given that, half-life $(\mathrm{T})=\frac{\ln 2}{\lambda}$ $\lambda=\frac{\ln 2}{\mathrm{~T}}$ We know that, law of radioactivity - $\mathrm{N}=\mathrm{N}_{\mathrm{o}}^{-\lambda \mathrm{t}}$ According to question - $\frac{\mathrm{N}_{\mathrm{o}}}{2}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ $\frac{1}{2}=\mathrm{e}^{-\lambda t_{1}}$ Taking $\log$ both the side, we get - $\lambda t_{1}=\ln 2$ Similarly, $\frac{\mathrm{N}_{\mathrm{o}}}{10}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$ $\frac{1}{10}=\mathrm{e}^{-\lambda \mathrm{t}_{2}}$ $\lambda \mathrm{t}_{2}=\ln 10$ Subtracting equation (ii) from equation (i), we get, $\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)=\ln 10-\ln 2$ $\frac{\ln 2}{\mathrm{~T}}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)=\ln 10-\ln 2$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\frac{\mathrm{T}(\ln 10-\ln 2)}{\ln 2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T}\left(\frac{\ln \left(\frac{10}{2}\right)}{\ln 2}\right)$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T} \frac{\ln 5}{\ln 2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T} \ln \left(\frac{5}{2}\right)$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T} \log \left(\frac{5}{2}\right)$
WB JEE 2013
NUCLEAR PHYSICS
147904
The half-life of radioactive material is $3 \mathrm{~h}$. If the initial amount is $300 \mathrm{~g}$. Then, after $18 \mathrm{~h}$, it will remain.
1 $4.69 \mathrm{~g}$
2 $46.8 \mathrm{~g}$
3 $9.375 \mathrm{~g}$
4 93.75
Explanation:
A Given that, Half-life of radioactive material $\left(\mathrm{T}_{1 / 2}\right)=3 \mathrm{~h}$ Initial amount of radioactive material $\left(\mathrm{N}_{\mathrm{O}}\right)=300 \mathrm{~g}$ Total time taken $(\mathrm{t})=18 \mathrm{~h}$ Amount of radioactive material left, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{18}{3}}$ $\mathrm{~N}=300 \times\left(\frac{1}{2}\right)^{6}$ $\mathrm{~N}=300 \times \frac{1}{64}$ $\mathrm{~N}=4.69 \mathrm{~g}$
147901
The half-life of ${ }^{60} \mathrm{Co}$ is approximately 5.25 years. In a sample containing $1 \mathrm{gm}$ of freshly prepared ${ }^{60} \mathrm{Co}$, how much of the isotope will be left after 21 years?
1 $125 \mathrm{mg}$
2 $62.5 \mathrm{mg}$
3 nothing will be left
4 $31.25 \mathrm{mg}$
Explanation:
B Given that, half-life of ${ }^{60} \mathrm{Co}\left(\mathrm{T}_{1 / 2}\right)=5.25 \mathrm{yr}$ Initial number of sample $\left(\mathrm{N}_{0}\right)=1 \mathrm{~g}$ Total time taken $(\mathrm{t})=21$ year We know that, Number of half-life $(\mathrm{n})=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{21}{5.25}=4$ Amount of sample left - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=1 \times\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}=\frac{1}{16} \mathrm{~g}=0.0625 \mathrm{~g}$ $\mathrm{~N}=62.5 \mathrm{mg}$
J and K CET-2012
NUCLEAR PHYSICS
147902
What amount of original radioactive material is left after 3 half-lives?
1 $6.5 \%$
2 $12.5 \%$
3 $25.5 \%$
4 $33.3 \%$
Explanation:
B Given that, Original amount of the radioactive material $=\mathrm{N}_{0}$ Number of half-life (n) $=3$ Amount of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{8}$ Hence, the percentage of original amount, $\mathrm{N}=\frac{\mathrm{N}_{0}}{8} \times 100$ $\mathrm{~N}=12.5 \% \text { of } \mathrm{N}_{0}$
J and K CET-2015
NUCLEAR PHYSICS
147903
The number of atoms of a radioactive substance of half-life $T$ is $N_{0}$ at $t=0$. The time necessary to decay from $N_{0} / 2$ atoms to $N_{0} / 10$ atoms will be
1 $\frac{5}{2} \mathrm{~T}$
2 $\mathrm{T} \log 5$
3 $\mathrm{T} \log \left[\frac{5}{2}\right]$
4 $\frac{T}{2} \log 5$
Explanation:
C Given that, half-life $(\mathrm{T})=\frac{\ln 2}{\lambda}$ $\lambda=\frac{\ln 2}{\mathrm{~T}}$ We know that, law of radioactivity - $\mathrm{N}=\mathrm{N}_{\mathrm{o}}^{-\lambda \mathrm{t}}$ According to question - $\frac{\mathrm{N}_{\mathrm{o}}}{2}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ $\frac{1}{2}=\mathrm{e}^{-\lambda t_{1}}$ Taking $\log$ both the side, we get - $\lambda t_{1}=\ln 2$ Similarly, $\frac{\mathrm{N}_{\mathrm{o}}}{10}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$ $\frac{1}{10}=\mathrm{e}^{-\lambda \mathrm{t}_{2}}$ $\lambda \mathrm{t}_{2}=\ln 10$ Subtracting equation (ii) from equation (i), we get, $\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)=\ln 10-\ln 2$ $\frac{\ln 2}{\mathrm{~T}}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)=\ln 10-\ln 2$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\frac{\mathrm{T}(\ln 10-\ln 2)}{\ln 2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T}\left(\frac{\ln \left(\frac{10}{2}\right)}{\ln 2}\right)$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T} \frac{\ln 5}{\ln 2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T} \ln \left(\frac{5}{2}\right)$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T} \log \left(\frac{5}{2}\right)$
WB JEE 2013
NUCLEAR PHYSICS
147904
The half-life of radioactive material is $3 \mathrm{~h}$. If the initial amount is $300 \mathrm{~g}$. Then, after $18 \mathrm{~h}$, it will remain.
1 $4.69 \mathrm{~g}$
2 $46.8 \mathrm{~g}$
3 $9.375 \mathrm{~g}$
4 93.75
Explanation:
A Given that, Half-life of radioactive material $\left(\mathrm{T}_{1 / 2}\right)=3 \mathrm{~h}$ Initial amount of radioactive material $\left(\mathrm{N}_{\mathrm{O}}\right)=300 \mathrm{~g}$ Total time taken $(\mathrm{t})=18 \mathrm{~h}$ Amount of radioactive material left, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{18}{3}}$ $\mathrm{~N}=300 \times\left(\frac{1}{2}\right)^{6}$ $\mathrm{~N}=300 \times \frac{1}{64}$ $\mathrm{~N}=4.69 \mathrm{~g}$
147901
The half-life of ${ }^{60} \mathrm{Co}$ is approximately 5.25 years. In a sample containing $1 \mathrm{gm}$ of freshly prepared ${ }^{60} \mathrm{Co}$, how much of the isotope will be left after 21 years?
1 $125 \mathrm{mg}$
2 $62.5 \mathrm{mg}$
3 nothing will be left
4 $31.25 \mathrm{mg}$
Explanation:
B Given that, half-life of ${ }^{60} \mathrm{Co}\left(\mathrm{T}_{1 / 2}\right)=5.25 \mathrm{yr}$ Initial number of sample $\left(\mathrm{N}_{0}\right)=1 \mathrm{~g}$ Total time taken $(\mathrm{t})=21$ year We know that, Number of half-life $(\mathrm{n})=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{21}{5.25}=4$ Amount of sample left - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=1 \times\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}=\frac{1}{16} \mathrm{~g}=0.0625 \mathrm{~g}$ $\mathrm{~N}=62.5 \mathrm{mg}$
J and K CET-2012
NUCLEAR PHYSICS
147902
What amount of original radioactive material is left after 3 half-lives?
1 $6.5 \%$
2 $12.5 \%$
3 $25.5 \%$
4 $33.3 \%$
Explanation:
B Given that, Original amount of the radioactive material $=\mathrm{N}_{0}$ Number of half-life (n) $=3$ Amount of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{8}$ Hence, the percentage of original amount, $\mathrm{N}=\frac{\mathrm{N}_{0}}{8} \times 100$ $\mathrm{~N}=12.5 \% \text { of } \mathrm{N}_{0}$
J and K CET-2015
NUCLEAR PHYSICS
147903
The number of atoms of a radioactive substance of half-life $T$ is $N_{0}$ at $t=0$. The time necessary to decay from $N_{0} / 2$ atoms to $N_{0} / 10$ atoms will be
1 $\frac{5}{2} \mathrm{~T}$
2 $\mathrm{T} \log 5$
3 $\mathrm{T} \log \left[\frac{5}{2}\right]$
4 $\frac{T}{2} \log 5$
Explanation:
C Given that, half-life $(\mathrm{T})=\frac{\ln 2}{\lambda}$ $\lambda=\frac{\ln 2}{\mathrm{~T}}$ We know that, law of radioactivity - $\mathrm{N}=\mathrm{N}_{\mathrm{o}}^{-\lambda \mathrm{t}}$ According to question - $\frac{\mathrm{N}_{\mathrm{o}}}{2}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ $\frac{1}{2}=\mathrm{e}^{-\lambda t_{1}}$ Taking $\log$ both the side, we get - $\lambda t_{1}=\ln 2$ Similarly, $\frac{\mathrm{N}_{\mathrm{o}}}{10}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$ $\frac{1}{10}=\mathrm{e}^{-\lambda \mathrm{t}_{2}}$ $\lambda \mathrm{t}_{2}=\ln 10$ Subtracting equation (ii) from equation (i), we get, $\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)=\ln 10-\ln 2$ $\frac{\ln 2}{\mathrm{~T}}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)=\ln 10-\ln 2$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\frac{\mathrm{T}(\ln 10-\ln 2)}{\ln 2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T}\left(\frac{\ln \left(\frac{10}{2}\right)}{\ln 2}\right)$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T} \frac{\ln 5}{\ln 2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T} \ln \left(\frac{5}{2}\right)$ $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{T} \log \left(\frac{5}{2}\right)$
WB JEE 2013
NUCLEAR PHYSICS
147904
The half-life of radioactive material is $3 \mathrm{~h}$. If the initial amount is $300 \mathrm{~g}$. Then, after $18 \mathrm{~h}$, it will remain.
1 $4.69 \mathrm{~g}$
2 $46.8 \mathrm{~g}$
3 $9.375 \mathrm{~g}$
4 93.75
Explanation:
A Given that, Half-life of radioactive material $\left(\mathrm{T}_{1 / 2}\right)=3 \mathrm{~h}$ Initial amount of radioactive material $\left(\mathrm{N}_{\mathrm{O}}\right)=300 \mathrm{~g}$ Total time taken $(\mathrm{t})=18 \mathrm{~h}$ Amount of radioactive material left, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{18}{3}}$ $\mathrm{~N}=300 \times\left(\frac{1}{2}\right)^{6}$ $\mathrm{~N}=300 \times \frac{1}{64}$ $\mathrm{~N}=4.69 \mathrm{~g}$
