364004
The radioactivity of a sample is \(R_{1}\) at a time \(T_{1}\) and \(R_{2}\) at a time \(T_{2}\). If the half-life of the specimen is \(T\), the number of atoms that have disintegrated in the time \(\left(T_{1}-T_{2}\right)\) is proportional to
1 \(\left(R_{1} T_{1}-R_{2} T_{2}\right)\)
2 \(\left(R_{1}-R_{2}\right)\)
3 \(\left(R_{1}-R_{2}\right) / T\)
4 \(\left(R_{1}-R_{2}\right) T\)
Explanation:
Given symbol \(T=T_{1 / 2}=\dfrac{0.693}{\lambda}\) \(\Rightarrow\) we can write \(\lambda=\left(\dfrac{0.693}{T}\right)\) \(\Rightarrow\) Radioactivity \(R=\lambda N\) \(R_{1}=\lambda N_{1}\) \(R_{2}=\lambda N_{2}\) (Since \(\lambda\) is same as only one sample is involved) Subtracting : \(\left(R_{1}-R_{2}\right)=\lambda\left(N_{1}-N_{2}\right)\) \(\Rightarrow\left(N_{1}-N_{2}\right)=\left(\dfrac{R_{1}-R_{2}}{\lambda}\right)\) \(=\left(\dfrac{R_{1}-R_{2}}{0.693}\right)(T)\) \(\Rightarrow\) Required \(\left(N_{1}-N_{2}\right) \propto\left[\left(R_{1}-R_{2}\right) T\right]\)
MHTCET - 2022
PHXII13:NUCLEI
364005
Assertion : Natural radioactivity was discovered by Henri Becquerel Reason : Becquerel is a unit of decay rate.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Actually decay rate shall have SI base unit \(\left(s^{-1}\right)\). Natural radioactivity was first identified in 1896 by French physicist Henri Becquerel. The unit of measurement for radioactivity, known as the becquerel, is defined as one disintegration per second, and it serves as one of the units for quantifying the rate of decay It is denoted by \(B q\). We have \(1 B q=2.7 \times 10^{-11} C i\). (where \(C i\) stands for Curie) So correct option is (2).
PHXII13:NUCLEI
364006
The half life of a radioactive substance is 20 minutes. The approximate time interval \(({t_2} - {t_1})\) between the time \({t_2}\) when \(\frac{2}{3}\) of it had decayed and time \({t_1}\) when \(\frac{1}{3}\) of it had decayed has
1 \(14\,\min \)
2 \(20\,\min \)
3 \(28\,\min \)
4 \(7\,\min \)
Explanation:
Number of undecayed nuclei after time \({t_2};\) \(\frac{{{N_0}}}{3} = {N_0}{e^{ - \lambda {t_2}}}\) (1) Number of undecayed nuclei after time \({t_1};\) \(\frac{{2{N_0}}}{3} = {N_0}{e^{ - \lambda {t_1}}}\) (2) From(1) \({e^{ - \lambda {t_2}}} = \frac{1}{3}\) \( \Rightarrow - \lambda {t_2} = {\log _e}\left( {\frac{1}{3}} \right)\) (3) From (2) \( - {e^{ - \lambda {t_1}}} = \frac{2}{3}\) \( \Rightarrow - \lambda {t_1} = {\log _e}\left( {\frac{2}{3}} \right)\) (4) Solving (3) and (4), we get \({t_2} - {t_1} = \frac{1}{\lambda }\left[ {{{\log }_e}\left( {\frac{2}{3}} \right) - {{\log }_e}\left( {\frac{1}{3}} \right)} \right] = \frac{{{{\log }_e}(2)}}{\lambda } = 20\min \)
PHXII13:NUCLEI
364007
1 curie represents
1 \(3.7 \times {10^7}\) disintegrations per second
2 \(3.7 \times {10^{10}}\) disintegrations per second
364004
The radioactivity of a sample is \(R_{1}\) at a time \(T_{1}\) and \(R_{2}\) at a time \(T_{2}\). If the half-life of the specimen is \(T\), the number of atoms that have disintegrated in the time \(\left(T_{1}-T_{2}\right)\) is proportional to
1 \(\left(R_{1} T_{1}-R_{2} T_{2}\right)\)
2 \(\left(R_{1}-R_{2}\right)\)
3 \(\left(R_{1}-R_{2}\right) / T\)
4 \(\left(R_{1}-R_{2}\right) T\)
Explanation:
Given symbol \(T=T_{1 / 2}=\dfrac{0.693}{\lambda}\) \(\Rightarrow\) we can write \(\lambda=\left(\dfrac{0.693}{T}\right)\) \(\Rightarrow\) Radioactivity \(R=\lambda N\) \(R_{1}=\lambda N_{1}\) \(R_{2}=\lambda N_{2}\) (Since \(\lambda\) is same as only one sample is involved) Subtracting : \(\left(R_{1}-R_{2}\right)=\lambda\left(N_{1}-N_{2}\right)\) \(\Rightarrow\left(N_{1}-N_{2}\right)=\left(\dfrac{R_{1}-R_{2}}{\lambda}\right)\) \(=\left(\dfrac{R_{1}-R_{2}}{0.693}\right)(T)\) \(\Rightarrow\) Required \(\left(N_{1}-N_{2}\right) \propto\left[\left(R_{1}-R_{2}\right) T\right]\)
MHTCET - 2022
PHXII13:NUCLEI
364005
Assertion : Natural radioactivity was discovered by Henri Becquerel Reason : Becquerel is a unit of decay rate.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Actually decay rate shall have SI base unit \(\left(s^{-1}\right)\). Natural radioactivity was first identified in 1896 by French physicist Henri Becquerel. The unit of measurement for radioactivity, known as the becquerel, is defined as one disintegration per second, and it serves as one of the units for quantifying the rate of decay It is denoted by \(B q\). We have \(1 B q=2.7 \times 10^{-11} C i\). (where \(C i\) stands for Curie) So correct option is (2).
PHXII13:NUCLEI
364006
The half life of a radioactive substance is 20 minutes. The approximate time interval \(({t_2} - {t_1})\) between the time \({t_2}\) when \(\frac{2}{3}\) of it had decayed and time \({t_1}\) when \(\frac{1}{3}\) of it had decayed has
1 \(14\,\min \)
2 \(20\,\min \)
3 \(28\,\min \)
4 \(7\,\min \)
Explanation:
Number of undecayed nuclei after time \({t_2};\) \(\frac{{{N_0}}}{3} = {N_0}{e^{ - \lambda {t_2}}}\) (1) Number of undecayed nuclei after time \({t_1};\) \(\frac{{2{N_0}}}{3} = {N_0}{e^{ - \lambda {t_1}}}\) (2) From(1) \({e^{ - \lambda {t_2}}} = \frac{1}{3}\) \( \Rightarrow - \lambda {t_2} = {\log _e}\left( {\frac{1}{3}} \right)\) (3) From (2) \( - {e^{ - \lambda {t_1}}} = \frac{2}{3}\) \( \Rightarrow - \lambda {t_1} = {\log _e}\left( {\frac{2}{3}} \right)\) (4) Solving (3) and (4), we get \({t_2} - {t_1} = \frac{1}{\lambda }\left[ {{{\log }_e}\left( {\frac{2}{3}} \right) - {{\log }_e}\left( {\frac{1}{3}} \right)} \right] = \frac{{{{\log }_e}(2)}}{\lambda } = 20\min \)
PHXII13:NUCLEI
364007
1 curie represents
1 \(3.7 \times {10^7}\) disintegrations per second
2 \(3.7 \times {10^{10}}\) disintegrations per second
364004
The radioactivity of a sample is \(R_{1}\) at a time \(T_{1}\) and \(R_{2}\) at a time \(T_{2}\). If the half-life of the specimen is \(T\), the number of atoms that have disintegrated in the time \(\left(T_{1}-T_{2}\right)\) is proportional to
1 \(\left(R_{1} T_{1}-R_{2} T_{2}\right)\)
2 \(\left(R_{1}-R_{2}\right)\)
3 \(\left(R_{1}-R_{2}\right) / T\)
4 \(\left(R_{1}-R_{2}\right) T\)
Explanation:
Given symbol \(T=T_{1 / 2}=\dfrac{0.693}{\lambda}\) \(\Rightarrow\) we can write \(\lambda=\left(\dfrac{0.693}{T}\right)\) \(\Rightarrow\) Radioactivity \(R=\lambda N\) \(R_{1}=\lambda N_{1}\) \(R_{2}=\lambda N_{2}\) (Since \(\lambda\) is same as only one sample is involved) Subtracting : \(\left(R_{1}-R_{2}\right)=\lambda\left(N_{1}-N_{2}\right)\) \(\Rightarrow\left(N_{1}-N_{2}\right)=\left(\dfrac{R_{1}-R_{2}}{\lambda}\right)\) \(=\left(\dfrac{R_{1}-R_{2}}{0.693}\right)(T)\) \(\Rightarrow\) Required \(\left(N_{1}-N_{2}\right) \propto\left[\left(R_{1}-R_{2}\right) T\right]\)
MHTCET - 2022
PHXII13:NUCLEI
364005
Assertion : Natural radioactivity was discovered by Henri Becquerel Reason : Becquerel is a unit of decay rate.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Actually decay rate shall have SI base unit \(\left(s^{-1}\right)\). Natural radioactivity was first identified in 1896 by French physicist Henri Becquerel. The unit of measurement for radioactivity, known as the becquerel, is defined as one disintegration per second, and it serves as one of the units for quantifying the rate of decay It is denoted by \(B q\). We have \(1 B q=2.7 \times 10^{-11} C i\). (where \(C i\) stands for Curie) So correct option is (2).
PHXII13:NUCLEI
364006
The half life of a radioactive substance is 20 minutes. The approximate time interval \(({t_2} - {t_1})\) between the time \({t_2}\) when \(\frac{2}{3}\) of it had decayed and time \({t_1}\) when \(\frac{1}{3}\) of it had decayed has
1 \(14\,\min \)
2 \(20\,\min \)
3 \(28\,\min \)
4 \(7\,\min \)
Explanation:
Number of undecayed nuclei after time \({t_2};\) \(\frac{{{N_0}}}{3} = {N_0}{e^{ - \lambda {t_2}}}\) (1) Number of undecayed nuclei after time \({t_1};\) \(\frac{{2{N_0}}}{3} = {N_0}{e^{ - \lambda {t_1}}}\) (2) From(1) \({e^{ - \lambda {t_2}}} = \frac{1}{3}\) \( \Rightarrow - \lambda {t_2} = {\log _e}\left( {\frac{1}{3}} \right)\) (3) From (2) \( - {e^{ - \lambda {t_1}}} = \frac{2}{3}\) \( \Rightarrow - \lambda {t_1} = {\log _e}\left( {\frac{2}{3}} \right)\) (4) Solving (3) and (4), we get \({t_2} - {t_1} = \frac{1}{\lambda }\left[ {{{\log }_e}\left( {\frac{2}{3}} \right) - {{\log }_e}\left( {\frac{1}{3}} \right)} \right] = \frac{{{{\log }_e}(2)}}{\lambda } = 20\min \)
PHXII13:NUCLEI
364007
1 curie represents
1 \(3.7 \times {10^7}\) disintegrations per second
2 \(3.7 \times {10^{10}}\) disintegrations per second
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII13:NUCLEI
364004
The radioactivity of a sample is \(R_{1}\) at a time \(T_{1}\) and \(R_{2}\) at a time \(T_{2}\). If the half-life of the specimen is \(T\), the number of atoms that have disintegrated in the time \(\left(T_{1}-T_{2}\right)\) is proportional to
1 \(\left(R_{1} T_{1}-R_{2} T_{2}\right)\)
2 \(\left(R_{1}-R_{2}\right)\)
3 \(\left(R_{1}-R_{2}\right) / T\)
4 \(\left(R_{1}-R_{2}\right) T\)
Explanation:
Given symbol \(T=T_{1 / 2}=\dfrac{0.693}{\lambda}\) \(\Rightarrow\) we can write \(\lambda=\left(\dfrac{0.693}{T}\right)\) \(\Rightarrow\) Radioactivity \(R=\lambda N\) \(R_{1}=\lambda N_{1}\) \(R_{2}=\lambda N_{2}\) (Since \(\lambda\) is same as only one sample is involved) Subtracting : \(\left(R_{1}-R_{2}\right)=\lambda\left(N_{1}-N_{2}\right)\) \(\Rightarrow\left(N_{1}-N_{2}\right)=\left(\dfrac{R_{1}-R_{2}}{\lambda}\right)\) \(=\left(\dfrac{R_{1}-R_{2}}{0.693}\right)(T)\) \(\Rightarrow\) Required \(\left(N_{1}-N_{2}\right) \propto\left[\left(R_{1}-R_{2}\right) T\right]\)
MHTCET - 2022
PHXII13:NUCLEI
364005
Assertion : Natural radioactivity was discovered by Henri Becquerel Reason : Becquerel is a unit of decay rate.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Actually decay rate shall have SI base unit \(\left(s^{-1}\right)\). Natural radioactivity was first identified in 1896 by French physicist Henri Becquerel. The unit of measurement for radioactivity, known as the becquerel, is defined as one disintegration per second, and it serves as one of the units for quantifying the rate of decay It is denoted by \(B q\). We have \(1 B q=2.7 \times 10^{-11} C i\). (where \(C i\) stands for Curie) So correct option is (2).
PHXII13:NUCLEI
364006
The half life of a radioactive substance is 20 minutes. The approximate time interval \(({t_2} - {t_1})\) between the time \({t_2}\) when \(\frac{2}{3}\) of it had decayed and time \({t_1}\) when \(\frac{1}{3}\) of it had decayed has
1 \(14\,\min \)
2 \(20\,\min \)
3 \(28\,\min \)
4 \(7\,\min \)
Explanation:
Number of undecayed nuclei after time \({t_2};\) \(\frac{{{N_0}}}{3} = {N_0}{e^{ - \lambda {t_2}}}\) (1) Number of undecayed nuclei after time \({t_1};\) \(\frac{{2{N_0}}}{3} = {N_0}{e^{ - \lambda {t_1}}}\) (2) From(1) \({e^{ - \lambda {t_2}}} = \frac{1}{3}\) \( \Rightarrow - \lambda {t_2} = {\log _e}\left( {\frac{1}{3}} \right)\) (3) From (2) \( - {e^{ - \lambda {t_1}}} = \frac{2}{3}\) \( \Rightarrow - \lambda {t_1} = {\log _e}\left( {\frac{2}{3}} \right)\) (4) Solving (3) and (4), we get \({t_2} - {t_1} = \frac{1}{\lambda }\left[ {{{\log }_e}\left( {\frac{2}{3}} \right) - {{\log }_e}\left( {\frac{1}{3}} \right)} \right] = \frac{{{{\log }_e}(2)}}{\lambda } = 20\min \)
PHXII13:NUCLEI
364007
1 curie represents
1 \(3.7 \times {10^7}\) disintegrations per second
2 \(3.7 \times {10^{10}}\) disintegrations per second
