363995
The half life of tritium is 12.5 years. What mass of tritium of initial mass 64 \(mg\) will remain undecayed after 50 years?
1 \(32\,mg\)
2 \(8\,mg\)
3 \(16\,mg\)
4 \(4\,mg\)
Explanation:
Here, Half life of tritium, \({T_{1/2}} = 12.5\) years Initial mass of tritium, \({m_0} = 64mg\) Time, \(t = 50\) years Mass of tritium remained undecayed after 50 years is, \(m = {m_0}{\left( {\frac{1}{2}} \right)^{t/{T_{1/2}}}} = \left( {64mg} \right){\left( {\frac{1}{2}} \right)^{50years/12.5years}}\) \( = \left( {64mg} \right){\left( {\frac{1}{2}} \right)^4} = \frac{{64mg}}{{16}} = 4mg\)
KCET - 2018
PHXII13:NUCLEI
363996
The half-life of a radioactive isotope \(X\) is 50 yr. It decays to another element Y which is stable. The two elements \(X\) and \(Y\) were found to be in the ratio of 1:15 in a sample of a given rock. The age of the rock was estimated to be
1 \(250\,yr\)
2 \(100\,yr\)
3 \(200\,yr\)
4 \(150\,yr\)
Explanation:
Let \({N_0}\) is the initial concentration of \(X\). The concentration of \(X\) at time \(t\) is \({N_X} = {N_0}{e^{ - \lambda t}}\) \({N_Y} = {N_0} - {N_0}{e^{ - \lambda t}}\) \(\frac{{{N_X}}}{{{N_Y}}} = \frac{{{N_0}{e^{ - \lambda t}}}}{{{N_0} - {N_0}{e^{ - \lambda t}}}} = \frac{1}{{15}}\) \( \Rightarrow 1 = 16{e^{ - \lambda t}} \Rightarrow t = \frac{{\ln 16}}{\lambda } = 200yr\)
PHXII13:NUCLEI
363997
Which one of the following nuclei has shorter mean life?
1 \(C\)
2 \(A\)
3 \({\text{Same for all}}\)
4 \(B\)
Explanation:
As \(\left| {\frac{{dN}}{{dt}}} \right|\) (activity) drops sharply in case of \(A\) so disintegration constant is higher for \(A\) Average life (Mean life) \( = \frac{1}{\lambda }\) So, average life will be shortest for \(A\)
PHXII13:NUCLEI
363998
The natural logarithm of the activity \({R}\) of a radioactive sample varies with time \({t}\) as shown. At \({t=0}\), there are \({N_{0}}\) undecayed nuclei. Then \({N_{0}}\) is equal to [Take \({e^{2}=7.5}\) ]
1 7,500
2 3,500
3 75,000
4 \({1,50,000}\)
Explanation:
From the given graph, we can write \({\log _{e} R_{0}=2}\) \({\Rightarrow R_{0}=e^{2}=7.5}\) \({R_{0}=\lambda N_{0}}\) As \({\log _{e} R=-\lambda t+\log _{e} R}\) \({\lambda=}\) slope \({=\dfrac{1}{10 \times 10^{3}}=\dfrac{1}{10^{4}}=10^{-4} / {sec}}\) \({N_{0}=\dfrac{R_{0}}{\lambda}=\dfrac{7.5}{10^{-4}}=75,000}\)
KCET - 2024
PHXII13:NUCLEI
363999
The graph shows the number of particles \({N_t}\) emitted per second by a radioactive source as a function of time \(t\). The relationship between \({N_t}\) and \(t\) is (take \(\ln \,20 = 3\))
1 \({N_t} = 1000{e^{ - 20t}}\)
2 \({N_t} = 20{e^{ - 20t}}\)
3 \({N_t} = 3{e^{ - 0.05t}}\)
4 \({N_t} = 20{e^{ - 0.05t}}\)
Explanation:
The graph can be represented by the linear equation as \(\frac{{\ln {N_t} - 3}}{{t - 0}} = \frac{{0 - 3}}{{60 - 0}}\) \( \Rightarrow \quad \ln {N_t} - \ln 20 = - \frac{1}{{20}}t\) \(\left( {{{\ln }_e}20 \approx 3} \right)\) \( \Rightarrow \quad \ln \frac{{{N_t}}}{{20}} = - 0.05t\) Hence,\({N_t} = 20{e^{ - 0.05t}}\)
363995
The half life of tritium is 12.5 years. What mass of tritium of initial mass 64 \(mg\) will remain undecayed after 50 years?
1 \(32\,mg\)
2 \(8\,mg\)
3 \(16\,mg\)
4 \(4\,mg\)
Explanation:
Here, Half life of tritium, \({T_{1/2}} = 12.5\) years Initial mass of tritium, \({m_0} = 64mg\) Time, \(t = 50\) years Mass of tritium remained undecayed after 50 years is, \(m = {m_0}{\left( {\frac{1}{2}} \right)^{t/{T_{1/2}}}} = \left( {64mg} \right){\left( {\frac{1}{2}} \right)^{50years/12.5years}}\) \( = \left( {64mg} \right){\left( {\frac{1}{2}} \right)^4} = \frac{{64mg}}{{16}} = 4mg\)
KCET - 2018
PHXII13:NUCLEI
363996
The half-life of a radioactive isotope \(X\) is 50 yr. It decays to another element Y which is stable. The two elements \(X\) and \(Y\) were found to be in the ratio of 1:15 in a sample of a given rock. The age of the rock was estimated to be
1 \(250\,yr\)
2 \(100\,yr\)
3 \(200\,yr\)
4 \(150\,yr\)
Explanation:
Let \({N_0}\) is the initial concentration of \(X\). The concentration of \(X\) at time \(t\) is \({N_X} = {N_0}{e^{ - \lambda t}}\) \({N_Y} = {N_0} - {N_0}{e^{ - \lambda t}}\) \(\frac{{{N_X}}}{{{N_Y}}} = \frac{{{N_0}{e^{ - \lambda t}}}}{{{N_0} - {N_0}{e^{ - \lambda t}}}} = \frac{1}{{15}}\) \( \Rightarrow 1 = 16{e^{ - \lambda t}} \Rightarrow t = \frac{{\ln 16}}{\lambda } = 200yr\)
PHXII13:NUCLEI
363997
Which one of the following nuclei has shorter mean life?
1 \(C\)
2 \(A\)
3 \({\text{Same for all}}\)
4 \(B\)
Explanation:
As \(\left| {\frac{{dN}}{{dt}}} \right|\) (activity) drops sharply in case of \(A\) so disintegration constant is higher for \(A\) Average life (Mean life) \( = \frac{1}{\lambda }\) So, average life will be shortest for \(A\)
PHXII13:NUCLEI
363998
The natural logarithm of the activity \({R}\) of a radioactive sample varies with time \({t}\) as shown. At \({t=0}\), there are \({N_{0}}\) undecayed nuclei. Then \({N_{0}}\) is equal to [Take \({e^{2}=7.5}\) ]
1 7,500
2 3,500
3 75,000
4 \({1,50,000}\)
Explanation:
From the given graph, we can write \({\log _{e} R_{0}=2}\) \({\Rightarrow R_{0}=e^{2}=7.5}\) \({R_{0}=\lambda N_{0}}\) As \({\log _{e} R=-\lambda t+\log _{e} R}\) \({\lambda=}\) slope \({=\dfrac{1}{10 \times 10^{3}}=\dfrac{1}{10^{4}}=10^{-4} / {sec}}\) \({N_{0}=\dfrac{R_{0}}{\lambda}=\dfrac{7.5}{10^{-4}}=75,000}\)
KCET - 2024
PHXII13:NUCLEI
363999
The graph shows the number of particles \({N_t}\) emitted per second by a radioactive source as a function of time \(t\). The relationship between \({N_t}\) and \(t\) is (take \(\ln \,20 = 3\))
1 \({N_t} = 1000{e^{ - 20t}}\)
2 \({N_t} = 20{e^{ - 20t}}\)
3 \({N_t} = 3{e^{ - 0.05t}}\)
4 \({N_t} = 20{e^{ - 0.05t}}\)
Explanation:
The graph can be represented by the linear equation as \(\frac{{\ln {N_t} - 3}}{{t - 0}} = \frac{{0 - 3}}{{60 - 0}}\) \( \Rightarrow \quad \ln {N_t} - \ln 20 = - \frac{1}{{20}}t\) \(\left( {{{\ln }_e}20 \approx 3} \right)\) \( \Rightarrow \quad \ln \frac{{{N_t}}}{{20}} = - 0.05t\) Hence,\({N_t} = 20{e^{ - 0.05t}}\)
363995
The half life of tritium is 12.5 years. What mass of tritium of initial mass 64 \(mg\) will remain undecayed after 50 years?
1 \(32\,mg\)
2 \(8\,mg\)
3 \(16\,mg\)
4 \(4\,mg\)
Explanation:
Here, Half life of tritium, \({T_{1/2}} = 12.5\) years Initial mass of tritium, \({m_0} = 64mg\) Time, \(t = 50\) years Mass of tritium remained undecayed after 50 years is, \(m = {m_0}{\left( {\frac{1}{2}} \right)^{t/{T_{1/2}}}} = \left( {64mg} \right){\left( {\frac{1}{2}} \right)^{50years/12.5years}}\) \( = \left( {64mg} \right){\left( {\frac{1}{2}} \right)^4} = \frac{{64mg}}{{16}} = 4mg\)
KCET - 2018
PHXII13:NUCLEI
363996
The half-life of a radioactive isotope \(X\) is 50 yr. It decays to another element Y which is stable. The two elements \(X\) and \(Y\) were found to be in the ratio of 1:15 in a sample of a given rock. The age of the rock was estimated to be
1 \(250\,yr\)
2 \(100\,yr\)
3 \(200\,yr\)
4 \(150\,yr\)
Explanation:
Let \({N_0}\) is the initial concentration of \(X\). The concentration of \(X\) at time \(t\) is \({N_X} = {N_0}{e^{ - \lambda t}}\) \({N_Y} = {N_0} - {N_0}{e^{ - \lambda t}}\) \(\frac{{{N_X}}}{{{N_Y}}} = \frac{{{N_0}{e^{ - \lambda t}}}}{{{N_0} - {N_0}{e^{ - \lambda t}}}} = \frac{1}{{15}}\) \( \Rightarrow 1 = 16{e^{ - \lambda t}} \Rightarrow t = \frac{{\ln 16}}{\lambda } = 200yr\)
PHXII13:NUCLEI
363997
Which one of the following nuclei has shorter mean life?
1 \(C\)
2 \(A\)
3 \({\text{Same for all}}\)
4 \(B\)
Explanation:
As \(\left| {\frac{{dN}}{{dt}}} \right|\) (activity) drops sharply in case of \(A\) so disintegration constant is higher for \(A\) Average life (Mean life) \( = \frac{1}{\lambda }\) So, average life will be shortest for \(A\)
PHXII13:NUCLEI
363998
The natural logarithm of the activity \({R}\) of a radioactive sample varies with time \({t}\) as shown. At \({t=0}\), there are \({N_{0}}\) undecayed nuclei. Then \({N_{0}}\) is equal to [Take \({e^{2}=7.5}\) ]
1 7,500
2 3,500
3 75,000
4 \({1,50,000}\)
Explanation:
From the given graph, we can write \({\log _{e} R_{0}=2}\) \({\Rightarrow R_{0}=e^{2}=7.5}\) \({R_{0}=\lambda N_{0}}\) As \({\log _{e} R=-\lambda t+\log _{e} R}\) \({\lambda=}\) slope \({=\dfrac{1}{10 \times 10^{3}}=\dfrac{1}{10^{4}}=10^{-4} / {sec}}\) \({N_{0}=\dfrac{R_{0}}{\lambda}=\dfrac{7.5}{10^{-4}}=75,000}\)
KCET - 2024
PHXII13:NUCLEI
363999
The graph shows the number of particles \({N_t}\) emitted per second by a radioactive source as a function of time \(t\). The relationship between \({N_t}\) and \(t\) is (take \(\ln \,20 = 3\))
1 \({N_t} = 1000{e^{ - 20t}}\)
2 \({N_t} = 20{e^{ - 20t}}\)
3 \({N_t} = 3{e^{ - 0.05t}}\)
4 \({N_t} = 20{e^{ - 0.05t}}\)
Explanation:
The graph can be represented by the linear equation as \(\frac{{\ln {N_t} - 3}}{{t - 0}} = \frac{{0 - 3}}{{60 - 0}}\) \( \Rightarrow \quad \ln {N_t} - \ln 20 = - \frac{1}{{20}}t\) \(\left( {{{\ln }_e}20 \approx 3} \right)\) \( \Rightarrow \quad \ln \frac{{{N_t}}}{{20}} = - 0.05t\) Hence,\({N_t} = 20{e^{ - 0.05t}}\)
363995
The half life of tritium is 12.5 years. What mass of tritium of initial mass 64 \(mg\) will remain undecayed after 50 years?
1 \(32\,mg\)
2 \(8\,mg\)
3 \(16\,mg\)
4 \(4\,mg\)
Explanation:
Here, Half life of tritium, \({T_{1/2}} = 12.5\) years Initial mass of tritium, \({m_0} = 64mg\) Time, \(t = 50\) years Mass of tritium remained undecayed after 50 years is, \(m = {m_0}{\left( {\frac{1}{2}} \right)^{t/{T_{1/2}}}} = \left( {64mg} \right){\left( {\frac{1}{2}} \right)^{50years/12.5years}}\) \( = \left( {64mg} \right){\left( {\frac{1}{2}} \right)^4} = \frac{{64mg}}{{16}} = 4mg\)
KCET - 2018
PHXII13:NUCLEI
363996
The half-life of a radioactive isotope \(X\) is 50 yr. It decays to another element Y which is stable. The two elements \(X\) and \(Y\) were found to be in the ratio of 1:15 in a sample of a given rock. The age of the rock was estimated to be
1 \(250\,yr\)
2 \(100\,yr\)
3 \(200\,yr\)
4 \(150\,yr\)
Explanation:
Let \({N_0}\) is the initial concentration of \(X\). The concentration of \(X\) at time \(t\) is \({N_X} = {N_0}{e^{ - \lambda t}}\) \({N_Y} = {N_0} - {N_0}{e^{ - \lambda t}}\) \(\frac{{{N_X}}}{{{N_Y}}} = \frac{{{N_0}{e^{ - \lambda t}}}}{{{N_0} - {N_0}{e^{ - \lambda t}}}} = \frac{1}{{15}}\) \( \Rightarrow 1 = 16{e^{ - \lambda t}} \Rightarrow t = \frac{{\ln 16}}{\lambda } = 200yr\)
PHXII13:NUCLEI
363997
Which one of the following nuclei has shorter mean life?
1 \(C\)
2 \(A\)
3 \({\text{Same for all}}\)
4 \(B\)
Explanation:
As \(\left| {\frac{{dN}}{{dt}}} \right|\) (activity) drops sharply in case of \(A\) so disintegration constant is higher for \(A\) Average life (Mean life) \( = \frac{1}{\lambda }\) So, average life will be shortest for \(A\)
PHXII13:NUCLEI
363998
The natural logarithm of the activity \({R}\) of a radioactive sample varies with time \({t}\) as shown. At \({t=0}\), there are \({N_{0}}\) undecayed nuclei. Then \({N_{0}}\) is equal to [Take \({e^{2}=7.5}\) ]
1 7,500
2 3,500
3 75,000
4 \({1,50,000}\)
Explanation:
From the given graph, we can write \({\log _{e} R_{0}=2}\) \({\Rightarrow R_{0}=e^{2}=7.5}\) \({R_{0}=\lambda N_{0}}\) As \({\log _{e} R=-\lambda t+\log _{e} R}\) \({\lambda=}\) slope \({=\dfrac{1}{10 \times 10^{3}}=\dfrac{1}{10^{4}}=10^{-4} / {sec}}\) \({N_{0}=\dfrac{R_{0}}{\lambda}=\dfrac{7.5}{10^{-4}}=75,000}\)
KCET - 2024
PHXII13:NUCLEI
363999
The graph shows the number of particles \({N_t}\) emitted per second by a radioactive source as a function of time \(t\). The relationship between \({N_t}\) and \(t\) is (take \(\ln \,20 = 3\))
1 \({N_t} = 1000{e^{ - 20t}}\)
2 \({N_t} = 20{e^{ - 20t}}\)
3 \({N_t} = 3{e^{ - 0.05t}}\)
4 \({N_t} = 20{e^{ - 0.05t}}\)
Explanation:
The graph can be represented by the linear equation as \(\frac{{\ln {N_t} - 3}}{{t - 0}} = \frac{{0 - 3}}{{60 - 0}}\) \( \Rightarrow \quad \ln {N_t} - \ln 20 = - \frac{1}{{20}}t\) \(\left( {{{\ln }_e}20 \approx 3} \right)\) \( \Rightarrow \quad \ln \frac{{{N_t}}}{{20}} = - 0.05t\) Hence,\({N_t} = 20{e^{ - 0.05t}}\)
363995
The half life of tritium is 12.5 years. What mass of tritium of initial mass 64 \(mg\) will remain undecayed after 50 years?
1 \(32\,mg\)
2 \(8\,mg\)
3 \(16\,mg\)
4 \(4\,mg\)
Explanation:
Here, Half life of tritium, \({T_{1/2}} = 12.5\) years Initial mass of tritium, \({m_0} = 64mg\) Time, \(t = 50\) years Mass of tritium remained undecayed after 50 years is, \(m = {m_0}{\left( {\frac{1}{2}} \right)^{t/{T_{1/2}}}} = \left( {64mg} \right){\left( {\frac{1}{2}} \right)^{50years/12.5years}}\) \( = \left( {64mg} \right){\left( {\frac{1}{2}} \right)^4} = \frac{{64mg}}{{16}} = 4mg\)
KCET - 2018
PHXII13:NUCLEI
363996
The half-life of a radioactive isotope \(X\) is 50 yr. It decays to another element Y which is stable. The two elements \(X\) and \(Y\) were found to be in the ratio of 1:15 in a sample of a given rock. The age of the rock was estimated to be
1 \(250\,yr\)
2 \(100\,yr\)
3 \(200\,yr\)
4 \(150\,yr\)
Explanation:
Let \({N_0}\) is the initial concentration of \(X\). The concentration of \(X\) at time \(t\) is \({N_X} = {N_0}{e^{ - \lambda t}}\) \({N_Y} = {N_0} - {N_0}{e^{ - \lambda t}}\) \(\frac{{{N_X}}}{{{N_Y}}} = \frac{{{N_0}{e^{ - \lambda t}}}}{{{N_0} - {N_0}{e^{ - \lambda t}}}} = \frac{1}{{15}}\) \( \Rightarrow 1 = 16{e^{ - \lambda t}} \Rightarrow t = \frac{{\ln 16}}{\lambda } = 200yr\)
PHXII13:NUCLEI
363997
Which one of the following nuclei has shorter mean life?
1 \(C\)
2 \(A\)
3 \({\text{Same for all}}\)
4 \(B\)
Explanation:
As \(\left| {\frac{{dN}}{{dt}}} \right|\) (activity) drops sharply in case of \(A\) so disintegration constant is higher for \(A\) Average life (Mean life) \( = \frac{1}{\lambda }\) So, average life will be shortest for \(A\)
PHXII13:NUCLEI
363998
The natural logarithm of the activity \({R}\) of a radioactive sample varies with time \({t}\) as shown. At \({t=0}\), there are \({N_{0}}\) undecayed nuclei. Then \({N_{0}}\) is equal to [Take \({e^{2}=7.5}\) ]
1 7,500
2 3,500
3 75,000
4 \({1,50,000}\)
Explanation:
From the given graph, we can write \({\log _{e} R_{0}=2}\) \({\Rightarrow R_{0}=e^{2}=7.5}\) \({R_{0}=\lambda N_{0}}\) As \({\log _{e} R=-\lambda t+\log _{e} R}\) \({\lambda=}\) slope \({=\dfrac{1}{10 \times 10^{3}}=\dfrac{1}{10^{4}}=10^{-4} / {sec}}\) \({N_{0}=\dfrac{R_{0}}{\lambda}=\dfrac{7.5}{10^{-4}}=75,000}\)
KCET - 2024
PHXII13:NUCLEI
363999
The graph shows the number of particles \({N_t}\) emitted per second by a radioactive source as a function of time \(t\). The relationship between \({N_t}\) and \(t\) is (take \(\ln \,20 = 3\))
1 \({N_t} = 1000{e^{ - 20t}}\)
2 \({N_t} = 20{e^{ - 20t}}\)
3 \({N_t} = 3{e^{ - 0.05t}}\)
4 \({N_t} = 20{e^{ - 0.05t}}\)
Explanation:
The graph can be represented by the linear equation as \(\frac{{\ln {N_t} - 3}}{{t - 0}} = \frac{{0 - 3}}{{60 - 0}}\) \( \Rightarrow \quad \ln {N_t} - \ln 20 = - \frac{1}{{20}}t\) \(\left( {{{\ln }_e}20 \approx 3} \right)\) \( \Rightarrow \quad \ln \frac{{{N_t}}}{{20}} = - 0.05t\) Hence,\({N_t} = 20{e^{ - 0.05t}}\)
