363654
Calculate binding energy of an alpha particle from the following data: Mass of free proton \( = 1.007825\,a.m.u\) Mass of free neutron \( = 1.008665\,a.m.u\) Mass of helium nucleus \( = 4.002800\,a.m.u\) \(1\,a.m.u = 931.5\,MeV\)
363655
The binding energies of the nuclei \(A\) and \(B\) are \(E_{a}\) and \(E_{b}\) respectively. Three nuclei of the element \(B\) fuse to give one nucleus of element ' \(A\) ' and an energy ' \(Q\) ' is released. Then \(E_{a}, E_{b}\), \(Q\) are related
1 \(E_{a}-3 E_{b}=Q\)
2 \(E_{b}-3 E_{a}=Q\)
3 \(E_{a}+3 E_{b}=Q\)
4 \(E_{b}+3 E_{a}=Q\)
Explanation:
Given reaction \(3 B \rightarrow A\) Energy released \(=B E\) of products \(-B E\) of reactants. \(\Rightarrow Q=E_{a}-3 E_{b}\)
PHXII13:NUCLEI
363656
Assertion : The binding energy per nucleon, for nuclei with atomic mass number \(A>100\), decrease with \(A\). Reason : The nuclear forces are weak for heavier nuclei.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect.
Explanation:
It is a well-established fact that the binding energy per nucleon decreases for heavy nuclei \((\mathrm{A}>100)\). This phenomenon is primarily due to the increasing influence of the Coulomb repulsion between protons as the number of protons in the nucleus increases. The Coulomb repulsion counteracts the attractive nuclear force, leading to a decrease in the binding energy per nucleon. The nuclear forces, often referred to as the strong nuclear force, are responsible for holding the nucleus together. The strong nuclear force is not weak for heavier nuclei; it is the dominant force that overcomes the electrostatic repulsion between protons and maintains nuclear stability. The decrease in binding energy per nucleon for heavy nuclei is primarily due to the increasing Coulomb repulsion between protons, not the weakness of the nuclear forces. The binding energy per nucleon decreases in heavy nuclei due to the Coulomb repulsion, not the weakness of the nuclear forces. So, the assertion is correct, but the reason provided is incorrect. So option (3) is correct
AIIMS - 2006
PHXII13:NUCLEI
363657
Binding energy per nucleon plot against the mass number for stable nuclei is shown in the figure. Which curve is correct?
1 \(A\)
2 \(B\)
3 \(C\)
4 \(D\)
Explanation:
The mass number increases, binding energy per nucleon increases up to \(\operatorname{Fe}(A=56)\). After atomic number greater than 56 , binding energy per nucleon decreases.
363654
Calculate binding energy of an alpha particle from the following data: Mass of free proton \( = 1.007825\,a.m.u\) Mass of free neutron \( = 1.008665\,a.m.u\) Mass of helium nucleus \( = 4.002800\,a.m.u\) \(1\,a.m.u = 931.5\,MeV\)
363655
The binding energies of the nuclei \(A\) and \(B\) are \(E_{a}\) and \(E_{b}\) respectively. Three nuclei of the element \(B\) fuse to give one nucleus of element ' \(A\) ' and an energy ' \(Q\) ' is released. Then \(E_{a}, E_{b}\), \(Q\) are related
1 \(E_{a}-3 E_{b}=Q\)
2 \(E_{b}-3 E_{a}=Q\)
3 \(E_{a}+3 E_{b}=Q\)
4 \(E_{b}+3 E_{a}=Q\)
Explanation:
Given reaction \(3 B \rightarrow A\) Energy released \(=B E\) of products \(-B E\) of reactants. \(\Rightarrow Q=E_{a}-3 E_{b}\)
PHXII13:NUCLEI
363656
Assertion : The binding energy per nucleon, for nuclei with atomic mass number \(A>100\), decrease with \(A\). Reason : The nuclear forces are weak for heavier nuclei.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect.
Explanation:
It is a well-established fact that the binding energy per nucleon decreases for heavy nuclei \((\mathrm{A}>100)\). This phenomenon is primarily due to the increasing influence of the Coulomb repulsion between protons as the number of protons in the nucleus increases. The Coulomb repulsion counteracts the attractive nuclear force, leading to a decrease in the binding energy per nucleon. The nuclear forces, often referred to as the strong nuclear force, are responsible for holding the nucleus together. The strong nuclear force is not weak for heavier nuclei; it is the dominant force that overcomes the electrostatic repulsion between protons and maintains nuclear stability. The decrease in binding energy per nucleon for heavy nuclei is primarily due to the increasing Coulomb repulsion between protons, not the weakness of the nuclear forces. The binding energy per nucleon decreases in heavy nuclei due to the Coulomb repulsion, not the weakness of the nuclear forces. So, the assertion is correct, but the reason provided is incorrect. So option (3) is correct
AIIMS - 2006
PHXII13:NUCLEI
363657
Binding energy per nucleon plot against the mass number for stable nuclei is shown in the figure. Which curve is correct?
1 \(A\)
2 \(B\)
3 \(C\)
4 \(D\)
Explanation:
The mass number increases, binding energy per nucleon increases up to \(\operatorname{Fe}(A=56)\). After atomic number greater than 56 , binding energy per nucleon decreases.
363654
Calculate binding energy of an alpha particle from the following data: Mass of free proton \( = 1.007825\,a.m.u\) Mass of free neutron \( = 1.008665\,a.m.u\) Mass of helium nucleus \( = 4.002800\,a.m.u\) \(1\,a.m.u = 931.5\,MeV\)
363655
The binding energies of the nuclei \(A\) and \(B\) are \(E_{a}\) and \(E_{b}\) respectively. Three nuclei of the element \(B\) fuse to give one nucleus of element ' \(A\) ' and an energy ' \(Q\) ' is released. Then \(E_{a}, E_{b}\), \(Q\) are related
1 \(E_{a}-3 E_{b}=Q\)
2 \(E_{b}-3 E_{a}=Q\)
3 \(E_{a}+3 E_{b}=Q\)
4 \(E_{b}+3 E_{a}=Q\)
Explanation:
Given reaction \(3 B \rightarrow A\) Energy released \(=B E\) of products \(-B E\) of reactants. \(\Rightarrow Q=E_{a}-3 E_{b}\)
PHXII13:NUCLEI
363656
Assertion : The binding energy per nucleon, for nuclei with atomic mass number \(A>100\), decrease with \(A\). Reason : The nuclear forces are weak for heavier nuclei.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect.
Explanation:
It is a well-established fact that the binding energy per nucleon decreases for heavy nuclei \((\mathrm{A}>100)\). This phenomenon is primarily due to the increasing influence of the Coulomb repulsion between protons as the number of protons in the nucleus increases. The Coulomb repulsion counteracts the attractive nuclear force, leading to a decrease in the binding energy per nucleon. The nuclear forces, often referred to as the strong nuclear force, are responsible for holding the nucleus together. The strong nuclear force is not weak for heavier nuclei; it is the dominant force that overcomes the electrostatic repulsion between protons and maintains nuclear stability. The decrease in binding energy per nucleon for heavy nuclei is primarily due to the increasing Coulomb repulsion between protons, not the weakness of the nuclear forces. The binding energy per nucleon decreases in heavy nuclei due to the Coulomb repulsion, not the weakness of the nuclear forces. So, the assertion is correct, but the reason provided is incorrect. So option (3) is correct
AIIMS - 2006
PHXII13:NUCLEI
363657
Binding energy per nucleon plot against the mass number for stable nuclei is shown in the figure. Which curve is correct?
1 \(A\)
2 \(B\)
3 \(C\)
4 \(D\)
Explanation:
The mass number increases, binding energy per nucleon increases up to \(\operatorname{Fe}(A=56)\). After atomic number greater than 56 , binding energy per nucleon decreases.
363654
Calculate binding energy of an alpha particle from the following data: Mass of free proton \( = 1.007825\,a.m.u\) Mass of free neutron \( = 1.008665\,a.m.u\) Mass of helium nucleus \( = 4.002800\,a.m.u\) \(1\,a.m.u = 931.5\,MeV\)
363655
The binding energies of the nuclei \(A\) and \(B\) are \(E_{a}\) and \(E_{b}\) respectively. Three nuclei of the element \(B\) fuse to give one nucleus of element ' \(A\) ' and an energy ' \(Q\) ' is released. Then \(E_{a}, E_{b}\), \(Q\) are related
1 \(E_{a}-3 E_{b}=Q\)
2 \(E_{b}-3 E_{a}=Q\)
3 \(E_{a}+3 E_{b}=Q\)
4 \(E_{b}+3 E_{a}=Q\)
Explanation:
Given reaction \(3 B \rightarrow A\) Energy released \(=B E\) of products \(-B E\) of reactants. \(\Rightarrow Q=E_{a}-3 E_{b}\)
PHXII13:NUCLEI
363656
Assertion : The binding energy per nucleon, for nuclei with atomic mass number \(A>100\), decrease with \(A\). Reason : The nuclear forces are weak for heavier nuclei.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect.
Explanation:
It is a well-established fact that the binding energy per nucleon decreases for heavy nuclei \((\mathrm{A}>100)\). This phenomenon is primarily due to the increasing influence of the Coulomb repulsion between protons as the number of protons in the nucleus increases. The Coulomb repulsion counteracts the attractive nuclear force, leading to a decrease in the binding energy per nucleon. The nuclear forces, often referred to as the strong nuclear force, are responsible for holding the nucleus together. The strong nuclear force is not weak for heavier nuclei; it is the dominant force that overcomes the electrostatic repulsion between protons and maintains nuclear stability. The decrease in binding energy per nucleon for heavy nuclei is primarily due to the increasing Coulomb repulsion between protons, not the weakness of the nuclear forces. The binding energy per nucleon decreases in heavy nuclei due to the Coulomb repulsion, not the weakness of the nuclear forces. So, the assertion is correct, but the reason provided is incorrect. So option (3) is correct
AIIMS - 2006
PHXII13:NUCLEI
363657
Binding energy per nucleon plot against the mass number for stable nuclei is shown in the figure. Which curve is correct?
1 \(A\)
2 \(B\)
3 \(C\)
4 \(D\)
Explanation:
The mass number increases, binding energy per nucleon increases up to \(\operatorname{Fe}(A=56)\). After atomic number greater than 56 , binding energy per nucleon decreases.
