363972
The half-life of radium is 1620 years and its atomic weight is 266. The number of atoms that will decay from its \(1g\) sample per second will be:
1 \(3.1 \times {10^{10}}\)
2 \(3.6 \times {10^{12}}\)
3 \(3.6 \times {10^{10}}\)
4 \(31.1 \times {10^{15}}\)
Explanation:
According to Avogadro’s hypothesis, \({N_0} = \frac{{6.02 \times {{10}^{23}}}}{{266}} = 2.66 \times {10^{21}}\) Half - time \( = T = \frac{{0.693}}{\lambda } = 1620\,{\rm{years}}\) \(\therefore \,\,\,\,\,\,\,\lambda = \frac{{0.693}}{{1620 \times 3.16 \times {{10}^7}}}\) \(\frac{{dN}}{{dt}} = \lambda {N_0} = 3.6 \times {10^{10}}\)
PHXII13:NUCLEI
363973
Two radioactive materials \({X_1}\) and \({X_2}\) have decay constants \(5\lambda \) and \(\lambda \) respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of \({X_1}\) to that of \({X_2}\) will be \(\frac{1}{e}\) after a time
1 \(\lambda \)
2 \(\frac{1}{2}\lambda \)
3 \(\frac{e}{\lambda }\)
4 \(\frac{1}{{4\lambda }}\)
Explanation:
The concentrations of \({X_1}\) and \({X_2}\) are \({N_1} = {N_0}{\mkern 1mu} {e^{ - 5\lambda t}}\,\,\,\,(1)\) \({N_2} = {N_0}{\mkern 1mu} {e^{ - \lambda t}}\,\,\,\,(2)\) Dividing Eq. (1) by Eq. (2), we obtain \(\frac{{{N_1}}}{{{N_2}}} = \frac{1}{e} = \frac{{{e^{ - 5\lambda t}}}}{{{e^{ - \lambda t}}}} = {e^{ - 4\lambda t}}\) \(\therefore \quad {e^{ - 1}} = {e^{ - 4\lambda t}} \Rightarrow 1 = 4\lambda t\) \( \Rightarrow t = \frac{1}{{4\lambda }}\)
363972
The half-life of radium is 1620 years and its atomic weight is 266. The number of atoms that will decay from its \(1g\) sample per second will be:
1 \(3.1 \times {10^{10}}\)
2 \(3.6 \times {10^{12}}\)
3 \(3.6 \times {10^{10}}\)
4 \(31.1 \times {10^{15}}\)
Explanation:
According to Avogadro’s hypothesis, \({N_0} = \frac{{6.02 \times {{10}^{23}}}}{{266}} = 2.66 \times {10^{21}}\) Half - time \( = T = \frac{{0.693}}{\lambda } = 1620\,{\rm{years}}\) \(\therefore \,\,\,\,\,\,\,\lambda = \frac{{0.693}}{{1620 \times 3.16 \times {{10}^7}}}\) \(\frac{{dN}}{{dt}} = \lambda {N_0} = 3.6 \times {10^{10}}\)
PHXII13:NUCLEI
363973
Two radioactive materials \({X_1}\) and \({X_2}\) have decay constants \(5\lambda \) and \(\lambda \) respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of \({X_1}\) to that of \({X_2}\) will be \(\frac{1}{e}\) after a time
1 \(\lambda \)
2 \(\frac{1}{2}\lambda \)
3 \(\frac{e}{\lambda }\)
4 \(\frac{1}{{4\lambda }}\)
Explanation:
The concentrations of \({X_1}\) and \({X_2}\) are \({N_1} = {N_0}{\mkern 1mu} {e^{ - 5\lambda t}}\,\,\,\,(1)\) \({N_2} = {N_0}{\mkern 1mu} {e^{ - \lambda t}}\,\,\,\,(2)\) Dividing Eq. (1) by Eq. (2), we obtain \(\frac{{{N_1}}}{{{N_2}}} = \frac{1}{e} = \frac{{{e^{ - 5\lambda t}}}}{{{e^{ - \lambda t}}}} = {e^{ - 4\lambda t}}\) \(\therefore \quad {e^{ - 1}} = {e^{ - 4\lambda t}} \Rightarrow 1 = 4\lambda t\) \( \Rightarrow t = \frac{1}{{4\lambda }}\)
363972
The half-life of radium is 1620 years and its atomic weight is 266. The number of atoms that will decay from its \(1g\) sample per second will be:
1 \(3.1 \times {10^{10}}\)
2 \(3.6 \times {10^{12}}\)
3 \(3.6 \times {10^{10}}\)
4 \(31.1 \times {10^{15}}\)
Explanation:
According to Avogadro’s hypothesis, \({N_0} = \frac{{6.02 \times {{10}^{23}}}}{{266}} = 2.66 \times {10^{21}}\) Half - time \( = T = \frac{{0.693}}{\lambda } = 1620\,{\rm{years}}\) \(\therefore \,\,\,\,\,\,\,\lambda = \frac{{0.693}}{{1620 \times 3.16 \times {{10}^7}}}\) \(\frac{{dN}}{{dt}} = \lambda {N_0} = 3.6 \times {10^{10}}\)
PHXII13:NUCLEI
363973
Two radioactive materials \({X_1}\) and \({X_2}\) have decay constants \(5\lambda \) and \(\lambda \) respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of \({X_1}\) to that of \({X_2}\) will be \(\frac{1}{e}\) after a time
1 \(\lambda \)
2 \(\frac{1}{2}\lambda \)
3 \(\frac{e}{\lambda }\)
4 \(\frac{1}{{4\lambda }}\)
Explanation:
The concentrations of \({X_1}\) and \({X_2}\) are \({N_1} = {N_0}{\mkern 1mu} {e^{ - 5\lambda t}}\,\,\,\,(1)\) \({N_2} = {N_0}{\mkern 1mu} {e^{ - \lambda t}}\,\,\,\,(2)\) Dividing Eq. (1) by Eq. (2), we obtain \(\frac{{{N_1}}}{{{N_2}}} = \frac{1}{e} = \frac{{{e^{ - 5\lambda t}}}}{{{e^{ - \lambda t}}}} = {e^{ - 4\lambda t}}\) \(\therefore \quad {e^{ - 1}} = {e^{ - 4\lambda t}} \Rightarrow 1 = 4\lambda t\) \( \Rightarrow t = \frac{1}{{4\lambda }}\)
363972
The half-life of radium is 1620 years and its atomic weight is 266. The number of atoms that will decay from its \(1g\) sample per second will be:
1 \(3.1 \times {10^{10}}\)
2 \(3.6 \times {10^{12}}\)
3 \(3.6 \times {10^{10}}\)
4 \(31.1 \times {10^{15}}\)
Explanation:
According to Avogadro’s hypothesis, \({N_0} = \frac{{6.02 \times {{10}^{23}}}}{{266}} = 2.66 \times {10^{21}}\) Half - time \( = T = \frac{{0.693}}{\lambda } = 1620\,{\rm{years}}\) \(\therefore \,\,\,\,\,\,\,\lambda = \frac{{0.693}}{{1620 \times 3.16 \times {{10}^7}}}\) \(\frac{{dN}}{{dt}} = \lambda {N_0} = 3.6 \times {10^{10}}\)
PHXII13:NUCLEI
363973
Two radioactive materials \({X_1}\) and \({X_2}\) have decay constants \(5\lambda \) and \(\lambda \) respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of \({X_1}\) to that of \({X_2}\) will be \(\frac{1}{e}\) after a time
1 \(\lambda \)
2 \(\frac{1}{2}\lambda \)
3 \(\frac{e}{\lambda }\)
4 \(\frac{1}{{4\lambda }}\)
Explanation:
The concentrations of \({X_1}\) and \({X_2}\) are \({N_1} = {N_0}{\mkern 1mu} {e^{ - 5\lambda t}}\,\,\,\,(1)\) \({N_2} = {N_0}{\mkern 1mu} {e^{ - \lambda t}}\,\,\,\,(2)\) Dividing Eq. (1) by Eq. (2), we obtain \(\frac{{{N_1}}}{{{N_2}}} = \frac{1}{e} = \frac{{{e^{ - 5\lambda t}}}}{{{e^{ - \lambda t}}}} = {e^{ - 4\lambda t}}\) \(\therefore \quad {e^{ - 1}} = {e^{ - 4\lambda t}} \Rightarrow 1 = 4\lambda t\) \( \Rightarrow t = \frac{1}{{4\lambda }}\)