147676
A radioactive sample contains $10^{-3} \mathrm{~kg}$ each of two nuclear species $A$ and $B$ with half-life 4 days and 8 days, respectively. The ratio of the amounts of $A$ and $B$ after period of 16 days is
1 $1: 2$
2 $4: 1$
3 $1: 4$
4 $2: 1$
5 $1: 1$
Explanation:
C Given, Half-life of species, $A=4$ days Half-life of species, $\mathrm{B}=8$ days For species $\mathrm{A}$, Number of half life $\left(n_{\mathrm{A}}\right)=\frac{\text { Time }(t)}{\text { Half life of species } A}$ $=\frac{16}{4}=4$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}_{\mathrm{A}}}$ Now, $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{4}$ Similarly for species B, $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{2}$ On dividing equation (i) by equation (ii), we get- $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{(2)^{2}}{(2)^{4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ $\mathrm{~N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
Kerala CEE - 2015
NUCLEAR PHYSICS
147677
the half-life of a radioactive substance is 20 min. The time taken between $50 \%$ decay and $87.5 \%$ decay of the substance will be
1 $20 \mathrm{~min}$
2 $30 \mathrm{~min}$
3 $40 \mathrm{~min}$
4 $25 \mathrm{~min}$
5 $10 \mathrm{~min}$
Explanation:
C Given, Half life of radioactive substance $=20 \mathrm{~min}$ So, Time taken to decay $50 \%$ is equal to half life of radioactive substance $=20 \mathrm{~min}$ $25 \%$ will decay in $=20 \mathrm{~min}$ Further $12.5 \%$ will decay in $=20 \mathrm{~min}$ $\therefore$ Total time of decay $=20+20+20=60 \mathrm{~min}$ So, difference in time the radioactive substance will decay $87.5 \%$ to $50 \%=60-20=40 \mathrm{~min}$
Kerala CEE- 2014
NUCLEAR PHYSICS
147678
A radioactive material of half-life time of 69.3 days kept in a container. $\frac{2}{3}$ rd of the substance remains undecayed after (given, In $\frac{3}{2}=0.4$ )
1 20 days
2 25 days
3 35 days
4 40 days
5 50 days
Explanation:
D Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=69.3$ days Total amount of substance $=\mathrm{N}_{0}$ Remaining substance undecayed $(\mathrm{N})=2 / 3 \mathrm{~N}_{0}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{e}^{\lambda \mathrm{t}}=\frac{\mathrm{N}_{0}}{\mathrm{~N}}$ Taking logarithm on both side, we get- $\lambda t=\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)$ $0.01 \times \mathrm{t}=\ln \left(\frac{\mathrm{N}_{0}}{(2 / 3) \mathrm{N}_{0}}\right)$ $\mathrm{t}=\frac{\ln (3 / 2)}{0.01}$ $\text { So, } \quad \mathrm{t}=\frac{0.4}{0.01}=40 \text { days }$
Kerala CEE- 2013
NUCLEAR PHYSICS
147680
A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After $5 \mathrm{~min}$, the rate becomes 1250 disintegration per minute. Then, its decay constant (per minute) is
1 $0.8 \log _{\mathrm{e}} 2$
2 $0.4 \log _{\mathrm{e}} 2$
3 $0.2 \log _{\mathrm{e}} 2$
4 $0.1 \log _{\mathrm{e}} 2$
5 $0.6 \log _{\mathrm{e}} 2$
Explanation:
B Given, Initial disintegration of sample $\left(\mathrm{N}_{0}\right)=5000$ Final disintegration of sample $(\mathrm{N})=1250$ Decay time $(\mathrm{t})=5 \mathrm{~min}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{e}^{\lambda \mathrm{t}}=\frac{\mathrm{N}_{0}}{\mathrm{~N}}$ Taking $\log$ in both sides, $\lambda=\frac{1}{\mathrm{t}} \ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)$ $\lambda=\frac{1}{\mathrm{t}} \ln \left(\frac{5000}{1250}\right)$ $\lambda=0.2 \log _{\mathrm{e}} 4$ $\lambda=0.4 \log _{\mathrm{e}} 2$
147676
A radioactive sample contains $10^{-3} \mathrm{~kg}$ each of two nuclear species $A$ and $B$ with half-life 4 days and 8 days, respectively. The ratio of the amounts of $A$ and $B$ after period of 16 days is
1 $1: 2$
2 $4: 1$
3 $1: 4$
4 $2: 1$
5 $1: 1$
Explanation:
C Given, Half-life of species, $A=4$ days Half-life of species, $\mathrm{B}=8$ days For species $\mathrm{A}$, Number of half life $\left(n_{\mathrm{A}}\right)=\frac{\text { Time }(t)}{\text { Half life of species } A}$ $=\frac{16}{4}=4$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}_{\mathrm{A}}}$ Now, $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{4}$ Similarly for species B, $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{2}$ On dividing equation (i) by equation (ii), we get- $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{(2)^{2}}{(2)^{4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ $\mathrm{~N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
Kerala CEE - 2015
NUCLEAR PHYSICS
147677
the half-life of a radioactive substance is 20 min. The time taken between $50 \%$ decay and $87.5 \%$ decay of the substance will be
1 $20 \mathrm{~min}$
2 $30 \mathrm{~min}$
3 $40 \mathrm{~min}$
4 $25 \mathrm{~min}$
5 $10 \mathrm{~min}$
Explanation:
C Given, Half life of radioactive substance $=20 \mathrm{~min}$ So, Time taken to decay $50 \%$ is equal to half life of radioactive substance $=20 \mathrm{~min}$ $25 \%$ will decay in $=20 \mathrm{~min}$ Further $12.5 \%$ will decay in $=20 \mathrm{~min}$ $\therefore$ Total time of decay $=20+20+20=60 \mathrm{~min}$ So, difference in time the radioactive substance will decay $87.5 \%$ to $50 \%=60-20=40 \mathrm{~min}$
Kerala CEE- 2014
NUCLEAR PHYSICS
147678
A radioactive material of half-life time of 69.3 days kept in a container. $\frac{2}{3}$ rd of the substance remains undecayed after (given, In $\frac{3}{2}=0.4$ )
1 20 days
2 25 days
3 35 days
4 40 days
5 50 days
Explanation:
D Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=69.3$ days Total amount of substance $=\mathrm{N}_{0}$ Remaining substance undecayed $(\mathrm{N})=2 / 3 \mathrm{~N}_{0}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{e}^{\lambda \mathrm{t}}=\frac{\mathrm{N}_{0}}{\mathrm{~N}}$ Taking logarithm on both side, we get- $\lambda t=\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)$ $0.01 \times \mathrm{t}=\ln \left(\frac{\mathrm{N}_{0}}{(2 / 3) \mathrm{N}_{0}}\right)$ $\mathrm{t}=\frac{\ln (3 / 2)}{0.01}$ $\text { So, } \quad \mathrm{t}=\frac{0.4}{0.01}=40 \text { days }$
Kerala CEE- 2013
NUCLEAR PHYSICS
147680
A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After $5 \mathrm{~min}$, the rate becomes 1250 disintegration per minute. Then, its decay constant (per minute) is
1 $0.8 \log _{\mathrm{e}} 2$
2 $0.4 \log _{\mathrm{e}} 2$
3 $0.2 \log _{\mathrm{e}} 2$
4 $0.1 \log _{\mathrm{e}} 2$
5 $0.6 \log _{\mathrm{e}} 2$
Explanation:
B Given, Initial disintegration of sample $\left(\mathrm{N}_{0}\right)=5000$ Final disintegration of sample $(\mathrm{N})=1250$ Decay time $(\mathrm{t})=5 \mathrm{~min}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{e}^{\lambda \mathrm{t}}=\frac{\mathrm{N}_{0}}{\mathrm{~N}}$ Taking $\log$ in both sides, $\lambda=\frac{1}{\mathrm{t}} \ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)$ $\lambda=\frac{1}{\mathrm{t}} \ln \left(\frac{5000}{1250}\right)$ $\lambda=0.2 \log _{\mathrm{e}} 4$ $\lambda=0.4 \log _{\mathrm{e}} 2$
147676
A radioactive sample contains $10^{-3} \mathrm{~kg}$ each of two nuclear species $A$ and $B$ with half-life 4 days and 8 days, respectively. The ratio of the amounts of $A$ and $B$ after period of 16 days is
1 $1: 2$
2 $4: 1$
3 $1: 4$
4 $2: 1$
5 $1: 1$
Explanation:
C Given, Half-life of species, $A=4$ days Half-life of species, $\mathrm{B}=8$ days For species $\mathrm{A}$, Number of half life $\left(n_{\mathrm{A}}\right)=\frac{\text { Time }(t)}{\text { Half life of species } A}$ $=\frac{16}{4}=4$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}_{\mathrm{A}}}$ Now, $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{4}$ Similarly for species B, $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{2}$ On dividing equation (i) by equation (ii), we get- $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{(2)^{2}}{(2)^{4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ $\mathrm{~N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
Kerala CEE - 2015
NUCLEAR PHYSICS
147677
the half-life of a radioactive substance is 20 min. The time taken between $50 \%$ decay and $87.5 \%$ decay of the substance will be
1 $20 \mathrm{~min}$
2 $30 \mathrm{~min}$
3 $40 \mathrm{~min}$
4 $25 \mathrm{~min}$
5 $10 \mathrm{~min}$
Explanation:
C Given, Half life of radioactive substance $=20 \mathrm{~min}$ So, Time taken to decay $50 \%$ is equal to half life of radioactive substance $=20 \mathrm{~min}$ $25 \%$ will decay in $=20 \mathrm{~min}$ Further $12.5 \%$ will decay in $=20 \mathrm{~min}$ $\therefore$ Total time of decay $=20+20+20=60 \mathrm{~min}$ So, difference in time the radioactive substance will decay $87.5 \%$ to $50 \%=60-20=40 \mathrm{~min}$
Kerala CEE- 2014
NUCLEAR PHYSICS
147678
A radioactive material of half-life time of 69.3 days kept in a container. $\frac{2}{3}$ rd of the substance remains undecayed after (given, In $\frac{3}{2}=0.4$ )
1 20 days
2 25 days
3 35 days
4 40 days
5 50 days
Explanation:
D Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=69.3$ days Total amount of substance $=\mathrm{N}_{0}$ Remaining substance undecayed $(\mathrm{N})=2 / 3 \mathrm{~N}_{0}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{e}^{\lambda \mathrm{t}}=\frac{\mathrm{N}_{0}}{\mathrm{~N}}$ Taking logarithm on both side, we get- $\lambda t=\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)$ $0.01 \times \mathrm{t}=\ln \left(\frac{\mathrm{N}_{0}}{(2 / 3) \mathrm{N}_{0}}\right)$ $\mathrm{t}=\frac{\ln (3 / 2)}{0.01}$ $\text { So, } \quad \mathrm{t}=\frac{0.4}{0.01}=40 \text { days }$
Kerala CEE- 2013
NUCLEAR PHYSICS
147680
A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After $5 \mathrm{~min}$, the rate becomes 1250 disintegration per minute. Then, its decay constant (per minute) is
1 $0.8 \log _{\mathrm{e}} 2$
2 $0.4 \log _{\mathrm{e}} 2$
3 $0.2 \log _{\mathrm{e}} 2$
4 $0.1 \log _{\mathrm{e}} 2$
5 $0.6 \log _{\mathrm{e}} 2$
Explanation:
B Given, Initial disintegration of sample $\left(\mathrm{N}_{0}\right)=5000$ Final disintegration of sample $(\mathrm{N})=1250$ Decay time $(\mathrm{t})=5 \mathrm{~min}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{e}^{\lambda \mathrm{t}}=\frac{\mathrm{N}_{0}}{\mathrm{~N}}$ Taking $\log$ in both sides, $\lambda=\frac{1}{\mathrm{t}} \ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)$ $\lambda=\frac{1}{\mathrm{t}} \ln \left(\frac{5000}{1250}\right)$ $\lambda=0.2 \log _{\mathrm{e}} 4$ $\lambda=0.4 \log _{\mathrm{e}} 2$
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NUCLEAR PHYSICS
147676
A radioactive sample contains $10^{-3} \mathrm{~kg}$ each of two nuclear species $A$ and $B$ with half-life 4 days and 8 days, respectively. The ratio of the amounts of $A$ and $B$ after period of 16 days is
1 $1: 2$
2 $4: 1$
3 $1: 4$
4 $2: 1$
5 $1: 1$
Explanation:
C Given, Half-life of species, $A=4$ days Half-life of species, $\mathrm{B}=8$ days For species $\mathrm{A}$, Number of half life $\left(n_{\mathrm{A}}\right)=\frac{\text { Time }(t)}{\text { Half life of species } A}$ $=\frac{16}{4}=4$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}_{\mathrm{A}}}$ Now, $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{4}$ Similarly for species B, $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{2}$ On dividing equation (i) by equation (ii), we get- $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{(2)^{2}}{(2)^{4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ $\mathrm{~N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
Kerala CEE - 2015
NUCLEAR PHYSICS
147677
the half-life of a radioactive substance is 20 min. The time taken between $50 \%$ decay and $87.5 \%$ decay of the substance will be
1 $20 \mathrm{~min}$
2 $30 \mathrm{~min}$
3 $40 \mathrm{~min}$
4 $25 \mathrm{~min}$
5 $10 \mathrm{~min}$
Explanation:
C Given, Half life of radioactive substance $=20 \mathrm{~min}$ So, Time taken to decay $50 \%$ is equal to half life of radioactive substance $=20 \mathrm{~min}$ $25 \%$ will decay in $=20 \mathrm{~min}$ Further $12.5 \%$ will decay in $=20 \mathrm{~min}$ $\therefore$ Total time of decay $=20+20+20=60 \mathrm{~min}$ So, difference in time the radioactive substance will decay $87.5 \%$ to $50 \%=60-20=40 \mathrm{~min}$
Kerala CEE- 2014
NUCLEAR PHYSICS
147678
A radioactive material of half-life time of 69.3 days kept in a container. $\frac{2}{3}$ rd of the substance remains undecayed after (given, In $\frac{3}{2}=0.4$ )
1 20 days
2 25 days
3 35 days
4 40 days
5 50 days
Explanation:
D Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=69.3$ days Total amount of substance $=\mathrm{N}_{0}$ Remaining substance undecayed $(\mathrm{N})=2 / 3 \mathrm{~N}_{0}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{e}^{\lambda \mathrm{t}}=\frac{\mathrm{N}_{0}}{\mathrm{~N}}$ Taking logarithm on both side, we get- $\lambda t=\ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)$ $0.01 \times \mathrm{t}=\ln \left(\frac{\mathrm{N}_{0}}{(2 / 3) \mathrm{N}_{0}}\right)$ $\mathrm{t}=\frac{\ln (3 / 2)}{0.01}$ $\text { So, } \quad \mathrm{t}=\frac{0.4}{0.01}=40 \text { days }$
Kerala CEE- 2013
NUCLEAR PHYSICS
147680
A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After $5 \mathrm{~min}$, the rate becomes 1250 disintegration per minute. Then, its decay constant (per minute) is
1 $0.8 \log _{\mathrm{e}} 2$
2 $0.4 \log _{\mathrm{e}} 2$
3 $0.2 \log _{\mathrm{e}} 2$
4 $0.1 \log _{\mathrm{e}} 2$
5 $0.6 \log _{\mathrm{e}} 2$
Explanation:
B Given, Initial disintegration of sample $\left(\mathrm{N}_{0}\right)=5000$ Final disintegration of sample $(\mathrm{N})=1250$ Decay time $(\mathrm{t})=5 \mathrm{~min}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{e}^{\lambda \mathrm{t}}=\frac{\mathrm{N}_{0}}{\mathrm{~N}}$ Taking $\log$ in both sides, $\lambda=\frac{1}{\mathrm{t}} \ln \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right)$ $\lambda=\frac{1}{\mathrm{t}} \ln \left(\frac{5000}{1250}\right)$ $\lambda=0.2 \log _{\mathrm{e}} 4$ $\lambda=0.4 \log _{\mathrm{e}} 2$
