363880
The fossil has a \({ }^{14} C:{ }^{12} C\) ratio, which is \(\left[\dfrac{1}{16}\right]\) of that in a living animal bone. If the half-life of \(^{14}C\) is \(5730\,yr\), then the age of the fossil bone is
1 \(11460\,yr\)
2 \(17190\,yr\)
3 \(22920\,yr\)
4 \(45840\,yr\)
Explanation:
After \(n\) half -lives (i.e. at \(t=n T_{\frac{1}{2}}\)), the number of nuclides left undecayed, \(N=N_{0}(1 / 2)^{n}\) Given, \(\dfrac{N}{N_{0}}=\dfrac{1}{16}\) \(\Rightarrow \dfrac{1}{16}=\left(\dfrac{1}{2}\right)^{n}\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{4}=\left(\dfrac{1}{2}\right)^{n}\) Equating the powers, we obtain, \(n=4\) i.e. \(\dfrac{t}{T_{\frac{1}{2}}}=4\) \(\Rightarrow t=4 T_{\frac{1}{2}}\) \(\therefore t = 4 \times 5730 = 22920\,yr\) \(\left( {\because {T_{\frac{1}{2}}} = 5730\,yr} \right)\)
AIIMS - 2006
PHXII13:NUCLEI
363881
A sample of radioactive material is used to provide desired doses of radiation for medical purposes. The total time for which the sample can be used will depend
1 Only on the number of times radiation is drawn from it
2 Only on the intensity of doses drawn from it
3 On both (1) and (2)
4 Neither on (1) nor on (2)
Explanation:
Radioactivity of a sample does not depend on how many times we use it. It is a continuous process and its rate does not depend on external factors.
PHXII13:NUCLEI
363882
Assertion : Cobalt -60 is useful in cancer therapy. Reason : Cobalt-60 is source of \(\gamma \)-radiations capable of killing cancerous cell.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Cobalt-60 is a source of \(\gamma \)-radiations. These radiation s kill cancer cells more rapidly than others because cancer cells are undifferentiated. However, healthy cells in vicinty of cancer cells are also damaged. Hence, option (1) is true.
PHXII13:NUCLEI
363883
Outside a nucleus,
1 Neutron is stable
2 Neutron is unstable
3 Proton and neutron both are stable
4 Proton and neutron both are unstable
Explanation:
The neutron has no net charge and mass is slightly larger than proton. Protons and neutrons constitute the nuclei of atoms. Outside the nucleus the neutron is unstable. The life of neutron outside the nucleus is about 932 seconds. ( \(\simeq 15.5\) minutes). After this period, \(n \rightarrow p+e^{-}+\bar{v}\)
363880
The fossil has a \({ }^{14} C:{ }^{12} C\) ratio, which is \(\left[\dfrac{1}{16}\right]\) of that in a living animal bone. If the half-life of \(^{14}C\) is \(5730\,yr\), then the age of the fossil bone is
1 \(11460\,yr\)
2 \(17190\,yr\)
3 \(22920\,yr\)
4 \(45840\,yr\)
Explanation:
After \(n\) half -lives (i.e. at \(t=n T_{\frac{1}{2}}\)), the number of nuclides left undecayed, \(N=N_{0}(1 / 2)^{n}\) Given, \(\dfrac{N}{N_{0}}=\dfrac{1}{16}\) \(\Rightarrow \dfrac{1}{16}=\left(\dfrac{1}{2}\right)^{n}\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{4}=\left(\dfrac{1}{2}\right)^{n}\) Equating the powers, we obtain, \(n=4\) i.e. \(\dfrac{t}{T_{\frac{1}{2}}}=4\) \(\Rightarrow t=4 T_{\frac{1}{2}}\) \(\therefore t = 4 \times 5730 = 22920\,yr\) \(\left( {\because {T_{\frac{1}{2}}} = 5730\,yr} \right)\)
AIIMS - 2006
PHXII13:NUCLEI
363881
A sample of radioactive material is used to provide desired doses of radiation for medical purposes. The total time for which the sample can be used will depend
1 Only on the number of times radiation is drawn from it
2 Only on the intensity of doses drawn from it
3 On both (1) and (2)
4 Neither on (1) nor on (2)
Explanation:
Radioactivity of a sample does not depend on how many times we use it. It is a continuous process and its rate does not depend on external factors.
PHXII13:NUCLEI
363882
Assertion : Cobalt -60 is useful in cancer therapy. Reason : Cobalt-60 is source of \(\gamma \)-radiations capable of killing cancerous cell.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Cobalt-60 is a source of \(\gamma \)-radiations. These radiation s kill cancer cells more rapidly than others because cancer cells are undifferentiated. However, healthy cells in vicinty of cancer cells are also damaged. Hence, option (1) is true.
PHXII13:NUCLEI
363883
Outside a nucleus,
1 Neutron is stable
2 Neutron is unstable
3 Proton and neutron both are stable
4 Proton and neutron both are unstable
Explanation:
The neutron has no net charge and mass is slightly larger than proton. Protons and neutrons constitute the nuclei of atoms. Outside the nucleus the neutron is unstable. The life of neutron outside the nucleus is about 932 seconds. ( \(\simeq 15.5\) minutes). After this period, \(n \rightarrow p+e^{-}+\bar{v}\)
363880
The fossil has a \({ }^{14} C:{ }^{12} C\) ratio, which is \(\left[\dfrac{1}{16}\right]\) of that in a living animal bone. If the half-life of \(^{14}C\) is \(5730\,yr\), then the age of the fossil bone is
1 \(11460\,yr\)
2 \(17190\,yr\)
3 \(22920\,yr\)
4 \(45840\,yr\)
Explanation:
After \(n\) half -lives (i.e. at \(t=n T_{\frac{1}{2}}\)), the number of nuclides left undecayed, \(N=N_{0}(1 / 2)^{n}\) Given, \(\dfrac{N}{N_{0}}=\dfrac{1}{16}\) \(\Rightarrow \dfrac{1}{16}=\left(\dfrac{1}{2}\right)^{n}\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{4}=\left(\dfrac{1}{2}\right)^{n}\) Equating the powers, we obtain, \(n=4\) i.e. \(\dfrac{t}{T_{\frac{1}{2}}}=4\) \(\Rightarrow t=4 T_{\frac{1}{2}}\) \(\therefore t = 4 \times 5730 = 22920\,yr\) \(\left( {\because {T_{\frac{1}{2}}} = 5730\,yr} \right)\)
AIIMS - 2006
PHXII13:NUCLEI
363881
A sample of radioactive material is used to provide desired doses of radiation for medical purposes. The total time for which the sample can be used will depend
1 Only on the number of times radiation is drawn from it
2 Only on the intensity of doses drawn from it
3 On both (1) and (2)
4 Neither on (1) nor on (2)
Explanation:
Radioactivity of a sample does not depend on how many times we use it. It is a continuous process and its rate does not depend on external factors.
PHXII13:NUCLEI
363882
Assertion : Cobalt -60 is useful in cancer therapy. Reason : Cobalt-60 is source of \(\gamma \)-radiations capable of killing cancerous cell.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Cobalt-60 is a source of \(\gamma \)-radiations. These radiation s kill cancer cells more rapidly than others because cancer cells are undifferentiated. However, healthy cells in vicinty of cancer cells are also damaged. Hence, option (1) is true.
PHXII13:NUCLEI
363883
Outside a nucleus,
1 Neutron is stable
2 Neutron is unstable
3 Proton and neutron both are stable
4 Proton and neutron both are unstable
Explanation:
The neutron has no net charge and mass is slightly larger than proton. Protons and neutrons constitute the nuclei of atoms. Outside the nucleus the neutron is unstable. The life of neutron outside the nucleus is about 932 seconds. ( \(\simeq 15.5\) minutes). After this period, \(n \rightarrow p+e^{-}+\bar{v}\)
363880
The fossil has a \({ }^{14} C:{ }^{12} C\) ratio, which is \(\left[\dfrac{1}{16}\right]\) of that in a living animal bone. If the half-life of \(^{14}C\) is \(5730\,yr\), then the age of the fossil bone is
1 \(11460\,yr\)
2 \(17190\,yr\)
3 \(22920\,yr\)
4 \(45840\,yr\)
Explanation:
After \(n\) half -lives (i.e. at \(t=n T_{\frac{1}{2}}\)), the number of nuclides left undecayed, \(N=N_{0}(1 / 2)^{n}\) Given, \(\dfrac{N}{N_{0}}=\dfrac{1}{16}\) \(\Rightarrow \dfrac{1}{16}=\left(\dfrac{1}{2}\right)^{n}\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{4}=\left(\dfrac{1}{2}\right)^{n}\) Equating the powers, we obtain, \(n=4\) i.e. \(\dfrac{t}{T_{\frac{1}{2}}}=4\) \(\Rightarrow t=4 T_{\frac{1}{2}}\) \(\therefore t = 4 \times 5730 = 22920\,yr\) \(\left( {\because {T_{\frac{1}{2}}} = 5730\,yr} \right)\)
AIIMS - 2006
PHXII13:NUCLEI
363881
A sample of radioactive material is used to provide desired doses of radiation for medical purposes. The total time for which the sample can be used will depend
1 Only on the number of times radiation is drawn from it
2 Only on the intensity of doses drawn from it
3 On both (1) and (2)
4 Neither on (1) nor on (2)
Explanation:
Radioactivity of a sample does not depend on how many times we use it. It is a continuous process and its rate does not depend on external factors.
PHXII13:NUCLEI
363882
Assertion : Cobalt -60 is useful in cancer therapy. Reason : Cobalt-60 is source of \(\gamma \)-radiations capable of killing cancerous cell.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Cobalt-60 is a source of \(\gamma \)-radiations. These radiation s kill cancer cells more rapidly than others because cancer cells are undifferentiated. However, healthy cells in vicinty of cancer cells are also damaged. Hence, option (1) is true.
PHXII13:NUCLEI
363883
Outside a nucleus,
1 Neutron is stable
2 Neutron is unstable
3 Proton and neutron both are stable
4 Proton and neutron both are unstable
Explanation:
The neutron has no net charge and mass is slightly larger than proton. Protons and neutrons constitute the nuclei of atoms. Outside the nucleus the neutron is unstable. The life of neutron outside the nucleus is about 932 seconds. ( \(\simeq 15.5\) minutes). After this period, \(n \rightarrow p+e^{-}+\bar{v}\)
