NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII13:NUCLEI
363721
If \(200\,MeV\) of energy is released in the fission of 1 nucleus of \(_{92}{U^{235}}\), then the number of nuclei that undergo fission to produce energy of \(10\,kWh\) in \(1s\), is:
363722
Most of the energy released in the fission is carried by
1 Neutrons
2 Fission fragments
3 Neutrons and fragments carry equally
4 Positrons
Explanation:
Conceptual Question
PHXII13:NUCLEI
363723
The energy released by the fission of one uranium atom is \(200\,MeV\). The number of fission per second required to produce \(3.2\,mW\) of power is: (take, \(1eV = 1.6 \times {10^{ - 19}}J\))
1 \({10^7}\)
2 \({10^{10}}\)
3 \({10^6}\)
4 \({10^8}\)
Explanation:
We have the energy released by fission of one uranium atom is \(200\,MeV\). So,\(E = 200 \times {10^6} \times 1.6 \times {10^{ - 19}} = 3.2 \times {10^{ - 11}}J\) The power required \( = 3.2 \times {10^{ - 3}}W\) Thus, the number of fission required is equal to \(\frac{{3.2 \times {{10}^{ - 3}}}}{{3.2 \times {{10}^{ - 11}}}} = {10^8}\) fissions.
PHXII13:NUCLEI
363724
The process of fission is responsible for the release of energy in
363721
If \(200\,MeV\) of energy is released in the fission of 1 nucleus of \(_{92}{U^{235}}\), then the number of nuclei that undergo fission to produce energy of \(10\,kWh\) in \(1s\), is:
363722
Most of the energy released in the fission is carried by
1 Neutrons
2 Fission fragments
3 Neutrons and fragments carry equally
4 Positrons
Explanation:
Conceptual Question
PHXII13:NUCLEI
363723
The energy released by the fission of one uranium atom is \(200\,MeV\). The number of fission per second required to produce \(3.2\,mW\) of power is: (take, \(1eV = 1.6 \times {10^{ - 19}}J\))
1 \({10^7}\)
2 \({10^{10}}\)
3 \({10^6}\)
4 \({10^8}\)
Explanation:
We have the energy released by fission of one uranium atom is \(200\,MeV\). So,\(E = 200 \times {10^6} \times 1.6 \times {10^{ - 19}} = 3.2 \times {10^{ - 11}}J\) The power required \( = 3.2 \times {10^{ - 3}}W\) Thus, the number of fission required is equal to \(\frac{{3.2 \times {{10}^{ - 3}}}}{{3.2 \times {{10}^{ - 11}}}} = {10^8}\) fissions.
PHXII13:NUCLEI
363724
The process of fission is responsible for the release of energy in
363721
If \(200\,MeV\) of energy is released in the fission of 1 nucleus of \(_{92}{U^{235}}\), then the number of nuclei that undergo fission to produce energy of \(10\,kWh\) in \(1s\), is:
363722
Most of the energy released in the fission is carried by
1 Neutrons
2 Fission fragments
3 Neutrons and fragments carry equally
4 Positrons
Explanation:
Conceptual Question
PHXII13:NUCLEI
363723
The energy released by the fission of one uranium atom is \(200\,MeV\). The number of fission per second required to produce \(3.2\,mW\) of power is: (take, \(1eV = 1.6 \times {10^{ - 19}}J\))
1 \({10^7}\)
2 \({10^{10}}\)
3 \({10^6}\)
4 \({10^8}\)
Explanation:
We have the energy released by fission of one uranium atom is \(200\,MeV\). So,\(E = 200 \times {10^6} \times 1.6 \times {10^{ - 19}} = 3.2 \times {10^{ - 11}}J\) The power required \( = 3.2 \times {10^{ - 3}}W\) Thus, the number of fission required is equal to \(\frac{{3.2 \times {{10}^{ - 3}}}}{{3.2 \times {{10}^{ - 11}}}} = {10^8}\) fissions.
PHXII13:NUCLEI
363724
The process of fission is responsible for the release of energy in
363721
If \(200\,MeV\) of energy is released in the fission of 1 nucleus of \(_{92}{U^{235}}\), then the number of nuclei that undergo fission to produce energy of \(10\,kWh\) in \(1s\), is:
363722
Most of the energy released in the fission is carried by
1 Neutrons
2 Fission fragments
3 Neutrons and fragments carry equally
4 Positrons
Explanation:
Conceptual Question
PHXII13:NUCLEI
363723
The energy released by the fission of one uranium atom is \(200\,MeV\). The number of fission per second required to produce \(3.2\,mW\) of power is: (take, \(1eV = 1.6 \times {10^{ - 19}}J\))
1 \({10^7}\)
2 \({10^{10}}\)
3 \({10^6}\)
4 \({10^8}\)
Explanation:
We have the energy released by fission of one uranium atom is \(200\,MeV\). So,\(E = 200 \times {10^6} \times 1.6 \times {10^{ - 19}} = 3.2 \times {10^{ - 11}}J\) The power required \( = 3.2 \times {10^{ - 3}}W\) Thus, the number of fission required is equal to \(\frac{{3.2 \times {{10}^{ - 3}}}}{{3.2 \times {{10}^{ - 11}}}} = {10^8}\) fissions.
PHXII13:NUCLEI
363724
The process of fission is responsible for the release of energy in
