363671
The binding energies of nuclei \(X\) and \(Y\) are \({E_1}\) and \({E_2}\) respecitvely. Two atoms of \(X\) fuse to give one atom of \(Y\) and an energy \(Q\) is released. Then
1 \(Q = 2{E_1} - {E_2}\)
2 \(Q = {E_2} - 2{E_1}\)
3 \(Q < 2{E_1} - {E_2}\)
4 \(Q > {E_2} - 2{E_1}\)
Explanation:
During fusion, binding energy of daughter nucleus is always greater than the total binding energy of the parent nuclei. The difference of binding energies is released. Hence , \(Q = {E_2} - 2{E_1}\)
PHXII13:NUCLEI
363672
Binding energy of a certain nucleus is \(18 \times {10^8}\;J.\) How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus?
363673
If \(M_{0}\) is the mass of isotope \({ }_{5}^{12} B, M_{p}\) and \(M_{n}\) are the masses of proton and neutron, then nuclear binding energy of isotope is
1 \(\left(5 M_{p}+7 M_{n}-M_{0}\right) c^{2}\)
2 \(\left(M_{0}-5 M_{p}\right) c^{2}\)
3 \(\left(M_{0}-12 M_{n}\right) c^{2}\)
4 \(\left(M_{0}-5 M_{p}-7 M_{n}\right) c^{2}\)
Explanation:
As we know, \(B.E. = \Delta \,m{c^2}\) Here \(\Delta m=Z M_{p}+(A-Z) M_{n}-M_{0}\) \(\Delta m=\left(5 M_{p}+7 M_{n}-M_{0}\right)\) So, \(B \cdot E .=\left(5 M_{p}+7 M_{n}-M_{0}\right) c^{2}\)
JEE - 2024
PHXII13:NUCLEI
363674
The mass defect of \({}_2^4He\) is 0.03\(u\). The binding energy per nucleon of helium(in \(MeV\)) is
1 \(69.825\)
2 \(6.9825\)
3 \(2.793\)
4 \(27.93\)
Explanation:
The mass defect of \({}_2^4He,\Delta m = 0.03u\) The binding energy per nucleon \( = \frac{{\Delta m \times 931}}{4}MeV\) \( = \frac{{0.03 \times 931}}{4}MeV = 6.9825MeV\)
363671
The binding energies of nuclei \(X\) and \(Y\) are \({E_1}\) and \({E_2}\) respecitvely. Two atoms of \(X\) fuse to give one atom of \(Y\) and an energy \(Q\) is released. Then
1 \(Q = 2{E_1} - {E_2}\)
2 \(Q = {E_2} - 2{E_1}\)
3 \(Q < 2{E_1} - {E_2}\)
4 \(Q > {E_2} - 2{E_1}\)
Explanation:
During fusion, binding energy of daughter nucleus is always greater than the total binding energy of the parent nuclei. The difference of binding energies is released. Hence , \(Q = {E_2} - 2{E_1}\)
PHXII13:NUCLEI
363672
Binding energy of a certain nucleus is \(18 \times {10^8}\;J.\) How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus?
363673
If \(M_{0}\) is the mass of isotope \({ }_{5}^{12} B, M_{p}\) and \(M_{n}\) are the masses of proton and neutron, then nuclear binding energy of isotope is
1 \(\left(5 M_{p}+7 M_{n}-M_{0}\right) c^{2}\)
2 \(\left(M_{0}-5 M_{p}\right) c^{2}\)
3 \(\left(M_{0}-12 M_{n}\right) c^{2}\)
4 \(\left(M_{0}-5 M_{p}-7 M_{n}\right) c^{2}\)
Explanation:
As we know, \(B.E. = \Delta \,m{c^2}\) Here \(\Delta m=Z M_{p}+(A-Z) M_{n}-M_{0}\) \(\Delta m=\left(5 M_{p}+7 M_{n}-M_{0}\right)\) So, \(B \cdot E .=\left(5 M_{p}+7 M_{n}-M_{0}\right) c^{2}\)
JEE - 2024
PHXII13:NUCLEI
363674
The mass defect of \({}_2^4He\) is 0.03\(u\). The binding energy per nucleon of helium(in \(MeV\)) is
1 \(69.825\)
2 \(6.9825\)
3 \(2.793\)
4 \(27.93\)
Explanation:
The mass defect of \({}_2^4He,\Delta m = 0.03u\) The binding energy per nucleon \( = \frac{{\Delta m \times 931}}{4}MeV\) \( = \frac{{0.03 \times 931}}{4}MeV = 6.9825MeV\)
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PHXII13:NUCLEI
363671
The binding energies of nuclei \(X\) and \(Y\) are \({E_1}\) and \({E_2}\) respecitvely. Two atoms of \(X\) fuse to give one atom of \(Y\) and an energy \(Q\) is released. Then
1 \(Q = 2{E_1} - {E_2}\)
2 \(Q = {E_2} - 2{E_1}\)
3 \(Q < 2{E_1} - {E_2}\)
4 \(Q > {E_2} - 2{E_1}\)
Explanation:
During fusion, binding energy of daughter nucleus is always greater than the total binding energy of the parent nuclei. The difference of binding energies is released. Hence , \(Q = {E_2} - 2{E_1}\)
PHXII13:NUCLEI
363672
Binding energy of a certain nucleus is \(18 \times {10^8}\;J.\) How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus?
363673
If \(M_{0}\) is the mass of isotope \({ }_{5}^{12} B, M_{p}\) and \(M_{n}\) are the masses of proton and neutron, then nuclear binding energy of isotope is
1 \(\left(5 M_{p}+7 M_{n}-M_{0}\right) c^{2}\)
2 \(\left(M_{0}-5 M_{p}\right) c^{2}\)
3 \(\left(M_{0}-12 M_{n}\right) c^{2}\)
4 \(\left(M_{0}-5 M_{p}-7 M_{n}\right) c^{2}\)
Explanation:
As we know, \(B.E. = \Delta \,m{c^2}\) Here \(\Delta m=Z M_{p}+(A-Z) M_{n}-M_{0}\) \(\Delta m=\left(5 M_{p}+7 M_{n}-M_{0}\right)\) So, \(B \cdot E .=\left(5 M_{p}+7 M_{n}-M_{0}\right) c^{2}\)
JEE - 2024
PHXII13:NUCLEI
363674
The mass defect of \({}_2^4He\) is 0.03\(u\). The binding energy per nucleon of helium(in \(MeV\)) is
1 \(69.825\)
2 \(6.9825\)
3 \(2.793\)
4 \(27.93\)
Explanation:
The mass defect of \({}_2^4He,\Delta m = 0.03u\) The binding energy per nucleon \( = \frac{{\Delta m \times 931}}{4}MeV\) \( = \frac{{0.03 \times 931}}{4}MeV = 6.9825MeV\)
363671
The binding energies of nuclei \(X\) and \(Y\) are \({E_1}\) and \({E_2}\) respecitvely. Two atoms of \(X\) fuse to give one atom of \(Y\) and an energy \(Q\) is released. Then
1 \(Q = 2{E_1} - {E_2}\)
2 \(Q = {E_2} - 2{E_1}\)
3 \(Q < 2{E_1} - {E_2}\)
4 \(Q > {E_2} - 2{E_1}\)
Explanation:
During fusion, binding energy of daughter nucleus is always greater than the total binding energy of the parent nuclei. The difference of binding energies is released. Hence , \(Q = {E_2} - 2{E_1}\)
PHXII13:NUCLEI
363672
Binding energy of a certain nucleus is \(18 \times {10^8}\;J.\) How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus?
363673
If \(M_{0}\) is the mass of isotope \({ }_{5}^{12} B, M_{p}\) and \(M_{n}\) are the masses of proton and neutron, then nuclear binding energy of isotope is
1 \(\left(5 M_{p}+7 M_{n}-M_{0}\right) c^{2}\)
2 \(\left(M_{0}-5 M_{p}\right) c^{2}\)
3 \(\left(M_{0}-12 M_{n}\right) c^{2}\)
4 \(\left(M_{0}-5 M_{p}-7 M_{n}\right) c^{2}\)
Explanation:
As we know, \(B.E. = \Delta \,m{c^2}\) Here \(\Delta m=Z M_{p}+(A-Z) M_{n}-M_{0}\) \(\Delta m=\left(5 M_{p}+7 M_{n}-M_{0}\right)\) So, \(B \cdot E .=\left(5 M_{p}+7 M_{n}-M_{0}\right) c^{2}\)
JEE - 2024
PHXII13:NUCLEI
363674
The mass defect of \({}_2^4He\) is 0.03\(u\). The binding energy per nucleon of helium(in \(MeV\)) is
1 \(69.825\)
2 \(6.9825\)
3 \(2.793\)
4 \(27.93\)
Explanation:
The mass defect of \({}_2^4He,\Delta m = 0.03u\) The binding energy per nucleon \( = \frac{{\Delta m \times 931}}{4}MeV\) \( = \frac{{0.03 \times 931}}{4}MeV = 6.9825MeV\)