363875
Carbon dating is best suited for determining the age of fossils if their age in years is of the order of
1 \({10^2}\)
2 \({10^4}\)
3 \({10^7}\)
4 \({10^6}\)
Explanation:
Carbon dating is best suited for determining the age of fossils if their age in years is of the order of 10,000 years. Beacause it is in the order of half life of \({C^{14}}\) which is 5730yr.
PHXII13:NUCLEI
363876
A bone fragment found in a cave contains 0.21 times as much \(_6^{14}C\) as an equal amount of carbon in air when the organism containing bone died. Find the approximate age of fragment \({t_{1/2}}\) of \(^{14}C = 5730\) years.
1 \(1.3 \times {10^4}y\)
2 \(1.15 \times {10^4}y\)
3 \(1.4 \times {10^4}y\)
4 \(1.24 \times {10^4}y\)
Explanation:
\(\lambda = \frac{{0.693}}{{{t_{t/2}}}} = \frac{{0.693}}{{5730}} = 1.209 \times {10^{ - 4}}{y^{ - 1}}\) Let \({m_0}\) is the initial mass of the \(\mathop C\limits^{14} .\) After time \(t\) the mass \(m = {m_0}{e^{ - \lambda t}}\) \(0.21\,{m_0} = {m_0}{e^{ - \lambda t}}\) \(0.21\, = {e^{ - \lambda t}} \Rightarrow \lambda = 1.3 \times {10^4}y\)
PHXII13:NUCLEI
363877
What is the age of an ancient wooden piece if it is known that the specific activity of \({C^{14}}\) nuclide in its amounts is \(3 / 5\) of that in freshly grown trees? Given the half life of \(C\) nuclide is \(5570 yr.\)
363878
Carbon 14 decays with half-life of about 5,800 years. In a sample of bone, the ratio of carbon 14 to carbon 12 is found to be \(\frac{1}{4}\) of what it is in free air. This bone may belong to a period about \(x\) centuries ago, where \(x\) is nearest to
1 \(58\)
2 \(2 \times 58\)
3 \(3 \times 58\)
4 \(58/2\)
Explanation:
To become \(\frac{1}{4}\)\(th\), it requires time of two half lives i.e., \(t = 2\left( {{T_{1/2}}} \right) = 2 \times 58\) Centuries
PHXII13:NUCLEI
363879
The fossil bone has a \(^{14}C{:^{12}}C\) ratio, which is \(\left( {\frac{1}{{16}}} \right)\) of that in a living animal bone. If the half-life time of \(^{14}C\) is 5730 years then the age of the fossil bone is
1 11460 years
2 17190 years
3 22920 years
4 45840 years
Explanation:
Let \({A_{14}}/{A_{12}}\) be the initial ratio of \(^{14}C{,^{12}}C\) . Let it be \(\frac{1}{{16}}{A_{14}}/{A_{12}}\) in the fossil. As \({A_{12}}\) remains constant, \({A_{14}}\) must have reduced by \(\frac{1}{{16}}\) during the period. No. of half lives required \( = 4\) Age of fossil \( = 4 \times 5730 = 22920\) years
363875
Carbon dating is best suited for determining the age of fossils if their age in years is of the order of
1 \({10^2}\)
2 \({10^4}\)
3 \({10^7}\)
4 \({10^6}\)
Explanation:
Carbon dating is best suited for determining the age of fossils if their age in years is of the order of 10,000 years. Beacause it is in the order of half life of \({C^{14}}\) which is 5730yr.
PHXII13:NUCLEI
363876
A bone fragment found in a cave contains 0.21 times as much \(_6^{14}C\) as an equal amount of carbon in air when the organism containing bone died. Find the approximate age of fragment \({t_{1/2}}\) of \(^{14}C = 5730\) years.
1 \(1.3 \times {10^4}y\)
2 \(1.15 \times {10^4}y\)
3 \(1.4 \times {10^4}y\)
4 \(1.24 \times {10^4}y\)
Explanation:
\(\lambda = \frac{{0.693}}{{{t_{t/2}}}} = \frac{{0.693}}{{5730}} = 1.209 \times {10^{ - 4}}{y^{ - 1}}\) Let \({m_0}\) is the initial mass of the \(\mathop C\limits^{14} .\) After time \(t\) the mass \(m = {m_0}{e^{ - \lambda t}}\) \(0.21\,{m_0} = {m_0}{e^{ - \lambda t}}\) \(0.21\, = {e^{ - \lambda t}} \Rightarrow \lambda = 1.3 \times {10^4}y\)
PHXII13:NUCLEI
363877
What is the age of an ancient wooden piece if it is known that the specific activity of \({C^{14}}\) nuclide in its amounts is \(3 / 5\) of that in freshly grown trees? Given the half life of \(C\) nuclide is \(5570 yr.\)
363878
Carbon 14 decays with half-life of about 5,800 years. In a sample of bone, the ratio of carbon 14 to carbon 12 is found to be \(\frac{1}{4}\) of what it is in free air. This bone may belong to a period about \(x\) centuries ago, where \(x\) is nearest to
1 \(58\)
2 \(2 \times 58\)
3 \(3 \times 58\)
4 \(58/2\)
Explanation:
To become \(\frac{1}{4}\)\(th\), it requires time of two half lives i.e., \(t = 2\left( {{T_{1/2}}} \right) = 2 \times 58\) Centuries
PHXII13:NUCLEI
363879
The fossil bone has a \(^{14}C{:^{12}}C\) ratio, which is \(\left( {\frac{1}{{16}}} \right)\) of that in a living animal bone. If the half-life time of \(^{14}C\) is 5730 years then the age of the fossil bone is
1 11460 years
2 17190 years
3 22920 years
4 45840 years
Explanation:
Let \({A_{14}}/{A_{12}}\) be the initial ratio of \(^{14}C{,^{12}}C\) . Let it be \(\frac{1}{{16}}{A_{14}}/{A_{12}}\) in the fossil. As \({A_{12}}\) remains constant, \({A_{14}}\) must have reduced by \(\frac{1}{{16}}\) during the period. No. of half lives required \( = 4\) Age of fossil \( = 4 \times 5730 = 22920\) years
363875
Carbon dating is best suited for determining the age of fossils if their age in years is of the order of
1 \({10^2}\)
2 \({10^4}\)
3 \({10^7}\)
4 \({10^6}\)
Explanation:
Carbon dating is best suited for determining the age of fossils if their age in years is of the order of 10,000 years. Beacause it is in the order of half life of \({C^{14}}\) which is 5730yr.
PHXII13:NUCLEI
363876
A bone fragment found in a cave contains 0.21 times as much \(_6^{14}C\) as an equal amount of carbon in air when the organism containing bone died. Find the approximate age of fragment \({t_{1/2}}\) of \(^{14}C = 5730\) years.
1 \(1.3 \times {10^4}y\)
2 \(1.15 \times {10^4}y\)
3 \(1.4 \times {10^4}y\)
4 \(1.24 \times {10^4}y\)
Explanation:
\(\lambda = \frac{{0.693}}{{{t_{t/2}}}} = \frac{{0.693}}{{5730}} = 1.209 \times {10^{ - 4}}{y^{ - 1}}\) Let \({m_0}\) is the initial mass of the \(\mathop C\limits^{14} .\) After time \(t\) the mass \(m = {m_0}{e^{ - \lambda t}}\) \(0.21\,{m_0} = {m_0}{e^{ - \lambda t}}\) \(0.21\, = {e^{ - \lambda t}} \Rightarrow \lambda = 1.3 \times {10^4}y\)
PHXII13:NUCLEI
363877
What is the age of an ancient wooden piece if it is known that the specific activity of \({C^{14}}\) nuclide in its amounts is \(3 / 5\) of that in freshly grown trees? Given the half life of \(C\) nuclide is \(5570 yr.\)
363878
Carbon 14 decays with half-life of about 5,800 years. In a sample of bone, the ratio of carbon 14 to carbon 12 is found to be \(\frac{1}{4}\) of what it is in free air. This bone may belong to a period about \(x\) centuries ago, where \(x\) is nearest to
1 \(58\)
2 \(2 \times 58\)
3 \(3 \times 58\)
4 \(58/2\)
Explanation:
To become \(\frac{1}{4}\)\(th\), it requires time of two half lives i.e., \(t = 2\left( {{T_{1/2}}} \right) = 2 \times 58\) Centuries
PHXII13:NUCLEI
363879
The fossil bone has a \(^{14}C{:^{12}}C\) ratio, which is \(\left( {\frac{1}{{16}}} \right)\) of that in a living animal bone. If the half-life time of \(^{14}C\) is 5730 years then the age of the fossil bone is
1 11460 years
2 17190 years
3 22920 years
4 45840 years
Explanation:
Let \({A_{14}}/{A_{12}}\) be the initial ratio of \(^{14}C{,^{12}}C\) . Let it be \(\frac{1}{{16}}{A_{14}}/{A_{12}}\) in the fossil. As \({A_{12}}\) remains constant, \({A_{14}}\) must have reduced by \(\frac{1}{{16}}\) during the period. No. of half lives required \( = 4\) Age of fossil \( = 4 \times 5730 = 22920\) years
363875
Carbon dating is best suited for determining the age of fossils if their age in years is of the order of
1 \({10^2}\)
2 \({10^4}\)
3 \({10^7}\)
4 \({10^6}\)
Explanation:
Carbon dating is best suited for determining the age of fossils if their age in years is of the order of 10,000 years. Beacause it is in the order of half life of \({C^{14}}\) which is 5730yr.
PHXII13:NUCLEI
363876
A bone fragment found in a cave contains 0.21 times as much \(_6^{14}C\) as an equal amount of carbon in air when the organism containing bone died. Find the approximate age of fragment \({t_{1/2}}\) of \(^{14}C = 5730\) years.
1 \(1.3 \times {10^4}y\)
2 \(1.15 \times {10^4}y\)
3 \(1.4 \times {10^4}y\)
4 \(1.24 \times {10^4}y\)
Explanation:
\(\lambda = \frac{{0.693}}{{{t_{t/2}}}} = \frac{{0.693}}{{5730}} = 1.209 \times {10^{ - 4}}{y^{ - 1}}\) Let \({m_0}\) is the initial mass of the \(\mathop C\limits^{14} .\) After time \(t\) the mass \(m = {m_0}{e^{ - \lambda t}}\) \(0.21\,{m_0} = {m_0}{e^{ - \lambda t}}\) \(0.21\, = {e^{ - \lambda t}} \Rightarrow \lambda = 1.3 \times {10^4}y\)
PHXII13:NUCLEI
363877
What is the age of an ancient wooden piece if it is known that the specific activity of \({C^{14}}\) nuclide in its amounts is \(3 / 5\) of that in freshly grown trees? Given the half life of \(C\) nuclide is \(5570 yr.\)
363878
Carbon 14 decays with half-life of about 5,800 years. In a sample of bone, the ratio of carbon 14 to carbon 12 is found to be \(\frac{1}{4}\) of what it is in free air. This bone may belong to a period about \(x\) centuries ago, where \(x\) is nearest to
1 \(58\)
2 \(2 \times 58\)
3 \(3 \times 58\)
4 \(58/2\)
Explanation:
To become \(\frac{1}{4}\)\(th\), it requires time of two half lives i.e., \(t = 2\left( {{T_{1/2}}} \right) = 2 \times 58\) Centuries
PHXII13:NUCLEI
363879
The fossil bone has a \(^{14}C{:^{12}}C\) ratio, which is \(\left( {\frac{1}{{16}}} \right)\) of that in a living animal bone. If the half-life time of \(^{14}C\) is 5730 years then the age of the fossil bone is
1 11460 years
2 17190 years
3 22920 years
4 45840 years
Explanation:
Let \({A_{14}}/{A_{12}}\) be the initial ratio of \(^{14}C{,^{12}}C\) . Let it be \(\frac{1}{{16}}{A_{14}}/{A_{12}}\) in the fossil. As \({A_{12}}\) remains constant, \({A_{14}}\) must have reduced by \(\frac{1}{{16}}\) during the period. No. of half lives required \( = 4\) Age of fossil \( = 4 \times 5730 = 22920\) years
363875
Carbon dating is best suited for determining the age of fossils if their age in years is of the order of
1 \({10^2}\)
2 \({10^4}\)
3 \({10^7}\)
4 \({10^6}\)
Explanation:
Carbon dating is best suited for determining the age of fossils if their age in years is of the order of 10,000 years. Beacause it is in the order of half life of \({C^{14}}\) which is 5730yr.
PHXII13:NUCLEI
363876
A bone fragment found in a cave contains 0.21 times as much \(_6^{14}C\) as an equal amount of carbon in air when the organism containing bone died. Find the approximate age of fragment \({t_{1/2}}\) of \(^{14}C = 5730\) years.
1 \(1.3 \times {10^4}y\)
2 \(1.15 \times {10^4}y\)
3 \(1.4 \times {10^4}y\)
4 \(1.24 \times {10^4}y\)
Explanation:
\(\lambda = \frac{{0.693}}{{{t_{t/2}}}} = \frac{{0.693}}{{5730}} = 1.209 \times {10^{ - 4}}{y^{ - 1}}\) Let \({m_0}\) is the initial mass of the \(\mathop C\limits^{14} .\) After time \(t\) the mass \(m = {m_0}{e^{ - \lambda t}}\) \(0.21\,{m_0} = {m_0}{e^{ - \lambda t}}\) \(0.21\, = {e^{ - \lambda t}} \Rightarrow \lambda = 1.3 \times {10^4}y\)
PHXII13:NUCLEI
363877
What is the age of an ancient wooden piece if it is known that the specific activity of \({C^{14}}\) nuclide in its amounts is \(3 / 5\) of that in freshly grown trees? Given the half life of \(C\) nuclide is \(5570 yr.\)
363878
Carbon 14 decays with half-life of about 5,800 years. In a sample of bone, the ratio of carbon 14 to carbon 12 is found to be \(\frac{1}{4}\) of what it is in free air. This bone may belong to a period about \(x\) centuries ago, where \(x\) is nearest to
1 \(58\)
2 \(2 \times 58\)
3 \(3 \times 58\)
4 \(58/2\)
Explanation:
To become \(\frac{1}{4}\)\(th\), it requires time of two half lives i.e., \(t = 2\left( {{T_{1/2}}} \right) = 2 \times 58\) Centuries
PHXII13:NUCLEI
363879
The fossil bone has a \(^{14}C{:^{12}}C\) ratio, which is \(\left( {\frac{1}{{16}}} \right)\) of that in a living animal bone. If the half-life time of \(^{14}C\) is 5730 years then the age of the fossil bone is
1 11460 years
2 17190 years
3 22920 years
4 45840 years
Explanation:
Let \({A_{14}}/{A_{12}}\) be the initial ratio of \(^{14}C{,^{12}}C\) . Let it be \(\frac{1}{{16}}{A_{14}}/{A_{12}}\) in the fossil. As \({A_{12}}\) remains constant, \({A_{14}}\) must have reduced by \(\frac{1}{{16}}\) during the period. No. of half lives required \( = 4\) Age of fossil \( = 4 \times 5730 = 22920\) years
