363619
The energy equivalent to a substance of mass 1\(g\) is
1 \(18 \times {10^{13}}J\)
2 \(9 \times {10^{13}}J\)
3 \(18 \times {10^6}J\)
4 \(9 \times {10^6}J\)
Explanation:
Here, Mass, \(m = 1,\,g = {10^{ - 3}}kg\) According to Einstein’s mass energy relation, the energy equivalent of mass \(m\) is \(E = m{c^2}\) ( where \(c\) is the speed of light in vaccum) \( = \left( {{{10}^{ - 3}}kg} \right){\left( {3 \times {{10}^8}m{s^{ - 1}}} \right)^2}\) \( = {10^{ - 3}} \times 9 \times {10^{16}}J = 9 \times {10^{13}}J\)
PHXII13:NUCLEI
363620
Let \({m_p}\) be the mass of a proton, \({m_n}\) the mass of a neturon, \({M_1}\) be the mass of \(_{10}N{e^{20}}\) nucleus Then
1 \({M_1} = 10\left( {{m_p} + {m_n}} \right)\)
2 \({M_1} > 10\left( {{m_p} + {m_n}} \right)\)
3 \({M_1} = 20{m_n}\)
4 \({M_1} < 10({m_p} + {m_n})\)
Explanation:
Since the mass of nucleus is less than that the sum of masses of nucleons forming the nucleus \(\therefore {M_1} < 10{m_p} + 10{m_n}\)
PHXII13:NUCLEI
363621
The atomic mass of \({ }_{6} C^{12}\) is \(12.000000{\rm{ }}\,\,u\) and that of \({ }_{6} C^{13}\) is \(13.003354{\rm{ }}\,\,u\). The required energy to remove a neutron from \({ }_{6} C^{13}\), if mass of neutron is \(1.008665{\rm{ }}\,\,u\), will be :
1 \(4.95\,MeV\)
2 \(6.25\,MeV\)
3 \(62.5\,MeV\)
4 \(49.5\,MeV\)
Explanation:
\({ }_{6}^{13} C \rightarrow{ }_{6}^{12} C+{ }_{0}^{1} n\) \(\Delta m = 12 + 1.008665 - 13.003354\) \(\Delta m = 0.005311\,\,amu\) Using mass energy relation i.e., \(u = \Delta m \times 931.5\,\,MeV\,\,\,\,\)(\(\Delta m = \) mass defect \()\) \(u=4.947\) \( \approx 4.95\,\,MeV\)
JEE - 2024
PHXII13:NUCLEI
363622
Energy equivalent to \(21\;g\) uranium is equal to:
363619
The energy equivalent to a substance of mass 1\(g\) is
1 \(18 \times {10^{13}}J\)
2 \(9 \times {10^{13}}J\)
3 \(18 \times {10^6}J\)
4 \(9 \times {10^6}J\)
Explanation:
Here, Mass, \(m = 1,\,g = {10^{ - 3}}kg\) According to Einstein’s mass energy relation, the energy equivalent of mass \(m\) is \(E = m{c^2}\) ( where \(c\) is the speed of light in vaccum) \( = \left( {{{10}^{ - 3}}kg} \right){\left( {3 \times {{10}^8}m{s^{ - 1}}} \right)^2}\) \( = {10^{ - 3}} \times 9 \times {10^{16}}J = 9 \times {10^{13}}J\)
PHXII13:NUCLEI
363620
Let \({m_p}\) be the mass of a proton, \({m_n}\) the mass of a neturon, \({M_1}\) be the mass of \(_{10}N{e^{20}}\) nucleus Then
1 \({M_1} = 10\left( {{m_p} + {m_n}} \right)\)
2 \({M_1} > 10\left( {{m_p} + {m_n}} \right)\)
3 \({M_1} = 20{m_n}\)
4 \({M_1} < 10({m_p} + {m_n})\)
Explanation:
Since the mass of nucleus is less than that the sum of masses of nucleons forming the nucleus \(\therefore {M_1} < 10{m_p} + 10{m_n}\)
PHXII13:NUCLEI
363621
The atomic mass of \({ }_{6} C^{12}\) is \(12.000000{\rm{ }}\,\,u\) and that of \({ }_{6} C^{13}\) is \(13.003354{\rm{ }}\,\,u\). The required energy to remove a neutron from \({ }_{6} C^{13}\), if mass of neutron is \(1.008665{\rm{ }}\,\,u\), will be :
1 \(4.95\,MeV\)
2 \(6.25\,MeV\)
3 \(62.5\,MeV\)
4 \(49.5\,MeV\)
Explanation:
\({ }_{6}^{13} C \rightarrow{ }_{6}^{12} C+{ }_{0}^{1} n\) \(\Delta m = 12 + 1.008665 - 13.003354\) \(\Delta m = 0.005311\,\,amu\) Using mass energy relation i.e., \(u = \Delta m \times 931.5\,\,MeV\,\,\,\,\)(\(\Delta m = \) mass defect \()\) \(u=4.947\) \( \approx 4.95\,\,MeV\)
JEE - 2024
PHXII13:NUCLEI
363622
Energy equivalent to \(21\;g\) uranium is equal to:
363619
The energy equivalent to a substance of mass 1\(g\) is
1 \(18 \times {10^{13}}J\)
2 \(9 \times {10^{13}}J\)
3 \(18 \times {10^6}J\)
4 \(9 \times {10^6}J\)
Explanation:
Here, Mass, \(m = 1,\,g = {10^{ - 3}}kg\) According to Einstein’s mass energy relation, the energy equivalent of mass \(m\) is \(E = m{c^2}\) ( where \(c\) is the speed of light in vaccum) \( = \left( {{{10}^{ - 3}}kg} \right){\left( {3 \times {{10}^8}m{s^{ - 1}}} \right)^2}\) \( = {10^{ - 3}} \times 9 \times {10^{16}}J = 9 \times {10^{13}}J\)
PHXII13:NUCLEI
363620
Let \({m_p}\) be the mass of a proton, \({m_n}\) the mass of a neturon, \({M_1}\) be the mass of \(_{10}N{e^{20}}\) nucleus Then
1 \({M_1} = 10\left( {{m_p} + {m_n}} \right)\)
2 \({M_1} > 10\left( {{m_p} + {m_n}} \right)\)
3 \({M_1} = 20{m_n}\)
4 \({M_1} < 10({m_p} + {m_n})\)
Explanation:
Since the mass of nucleus is less than that the sum of masses of nucleons forming the nucleus \(\therefore {M_1} < 10{m_p} + 10{m_n}\)
PHXII13:NUCLEI
363621
The atomic mass of \({ }_{6} C^{12}\) is \(12.000000{\rm{ }}\,\,u\) and that of \({ }_{6} C^{13}\) is \(13.003354{\rm{ }}\,\,u\). The required energy to remove a neutron from \({ }_{6} C^{13}\), if mass of neutron is \(1.008665{\rm{ }}\,\,u\), will be :
1 \(4.95\,MeV\)
2 \(6.25\,MeV\)
3 \(62.5\,MeV\)
4 \(49.5\,MeV\)
Explanation:
\({ }_{6}^{13} C \rightarrow{ }_{6}^{12} C+{ }_{0}^{1} n\) \(\Delta m = 12 + 1.008665 - 13.003354\) \(\Delta m = 0.005311\,\,amu\) Using mass energy relation i.e., \(u = \Delta m \times 931.5\,\,MeV\,\,\,\,\)(\(\Delta m = \) mass defect \()\) \(u=4.947\) \( \approx 4.95\,\,MeV\)
JEE - 2024
PHXII13:NUCLEI
363622
Energy equivalent to \(21\;g\) uranium is equal to:
363619
The energy equivalent to a substance of mass 1\(g\) is
1 \(18 \times {10^{13}}J\)
2 \(9 \times {10^{13}}J\)
3 \(18 \times {10^6}J\)
4 \(9 \times {10^6}J\)
Explanation:
Here, Mass, \(m = 1,\,g = {10^{ - 3}}kg\) According to Einstein’s mass energy relation, the energy equivalent of mass \(m\) is \(E = m{c^2}\) ( where \(c\) is the speed of light in vaccum) \( = \left( {{{10}^{ - 3}}kg} \right){\left( {3 \times {{10}^8}m{s^{ - 1}}} \right)^2}\) \( = {10^{ - 3}} \times 9 \times {10^{16}}J = 9 \times {10^{13}}J\)
PHXII13:NUCLEI
363620
Let \({m_p}\) be the mass of a proton, \({m_n}\) the mass of a neturon, \({M_1}\) be the mass of \(_{10}N{e^{20}}\) nucleus Then
1 \({M_1} = 10\left( {{m_p} + {m_n}} \right)\)
2 \({M_1} > 10\left( {{m_p} + {m_n}} \right)\)
3 \({M_1} = 20{m_n}\)
4 \({M_1} < 10({m_p} + {m_n})\)
Explanation:
Since the mass of nucleus is less than that the sum of masses of nucleons forming the nucleus \(\therefore {M_1} < 10{m_p} + 10{m_n}\)
PHXII13:NUCLEI
363621
The atomic mass of \({ }_{6} C^{12}\) is \(12.000000{\rm{ }}\,\,u\) and that of \({ }_{6} C^{13}\) is \(13.003354{\rm{ }}\,\,u\). The required energy to remove a neutron from \({ }_{6} C^{13}\), if mass of neutron is \(1.008665{\rm{ }}\,\,u\), will be :
1 \(4.95\,MeV\)
2 \(6.25\,MeV\)
3 \(62.5\,MeV\)
4 \(49.5\,MeV\)
Explanation:
\({ }_{6}^{13} C \rightarrow{ }_{6}^{12} C+{ }_{0}^{1} n\) \(\Delta m = 12 + 1.008665 - 13.003354\) \(\Delta m = 0.005311\,\,amu\) Using mass energy relation i.e., \(u = \Delta m \times 931.5\,\,MeV\,\,\,\,\)(\(\Delta m = \) mass defect \()\) \(u=4.947\) \( \approx 4.95\,\,MeV\)
JEE - 2024
PHXII13:NUCLEI
363622
Energy equivalent to \(21\;g\) uranium is equal to:
363619
The energy equivalent to a substance of mass 1\(g\) is
1 \(18 \times {10^{13}}J\)
2 \(9 \times {10^{13}}J\)
3 \(18 \times {10^6}J\)
4 \(9 \times {10^6}J\)
Explanation:
Here, Mass, \(m = 1,\,g = {10^{ - 3}}kg\) According to Einstein’s mass energy relation, the energy equivalent of mass \(m\) is \(E = m{c^2}\) ( where \(c\) is the speed of light in vaccum) \( = \left( {{{10}^{ - 3}}kg} \right){\left( {3 \times {{10}^8}m{s^{ - 1}}} \right)^2}\) \( = {10^{ - 3}} \times 9 \times {10^{16}}J = 9 \times {10^{13}}J\)
PHXII13:NUCLEI
363620
Let \({m_p}\) be the mass of a proton, \({m_n}\) the mass of a neturon, \({M_1}\) be the mass of \(_{10}N{e^{20}}\) nucleus Then
1 \({M_1} = 10\left( {{m_p} + {m_n}} \right)\)
2 \({M_1} > 10\left( {{m_p} + {m_n}} \right)\)
3 \({M_1} = 20{m_n}\)
4 \({M_1} < 10({m_p} + {m_n})\)
Explanation:
Since the mass of nucleus is less than that the sum of masses of nucleons forming the nucleus \(\therefore {M_1} < 10{m_p} + 10{m_n}\)
PHXII13:NUCLEI
363621
The atomic mass of \({ }_{6} C^{12}\) is \(12.000000{\rm{ }}\,\,u\) and that of \({ }_{6} C^{13}\) is \(13.003354{\rm{ }}\,\,u\). The required energy to remove a neutron from \({ }_{6} C^{13}\), if mass of neutron is \(1.008665{\rm{ }}\,\,u\), will be :
1 \(4.95\,MeV\)
2 \(6.25\,MeV\)
3 \(62.5\,MeV\)
4 \(49.5\,MeV\)
Explanation:
\({ }_{6}^{13} C \rightarrow{ }_{6}^{12} C+{ }_{0}^{1} n\) \(\Delta m = 12 + 1.008665 - 13.003354\) \(\Delta m = 0.005311\,\,amu\) Using mass energy relation i.e., \(u = \Delta m \times 931.5\,\,MeV\,\,\,\,\)(\(\Delta m = \) mass defect \()\) \(u=4.947\) \( \approx 4.95\,\,MeV\)
JEE - 2024
PHXII13:NUCLEI
363622
Energy equivalent to \(21\;g\) uranium is equal to:
