
${{\text{M}}_{\text{p}}}$ denotes the mass of a proton and ${{\text{M}}_{\text{n}}}$that of a neutron. A given nucleus, of binding energy ${\text{B}}$, contains ${\text{Z}}$ protons and ${\text{N}}$ neutrons. The mass \[{\text{M}}\left( {{\text{N,Z}}} \right){\text{}}\] of the nucleus is given by (c is the velocity of light)
A.\[{\text{M}}\left( {{\text{N,Z}}} \right){\text{ = N}}{{\text{M}}_{\text{n}}}{\text{ + Z}}{{\text{M}}_{\text{p}}}{\text{ + B}}{{\text{c}}^{\text{2}}}\]
B.\[{\text{M}}\left( {{\text{N,Z}}} \right){\text{ = N}}{{\text{M}}_{\text{n}}}{\text{ + Z}}{{\text{M}}_{\text{p}}}{\text{ - }}\dfrac{{\text{B}}}{{{{\text{c}}^{\text{2}}}}}{\text{ }}\]
C.\[{\text{M}}\left( {{\text{N,Z}}} \right){\text{ = N}}{{\text{M}}_{\text{n}}}{\text{+ Z}}{{\text{M}}_{\text{p}}}{\text{ + }}\dfrac{{\text{B}}}{{{{\text{c}}^{\text{2}}}}}\]
D.\[{\text{M}}\left( {{\text{N,Z}}} \right){\text{ = N}}{{\text{M}}_{\text{n}}}{\text{ + Z}}{{\text{M}}_{\text{p}}}{\text{ - B}}{{\text{c}}^{\text{2}}}\]
Answer
583.2k+ views
Hint: To answer this question, recall the concept of Einstein's theory of relativity. We shall find the mass defect in terms of the binding energy of an atom. Substitute both the equations and rearrange the terms in the formula to answer this question.
Formula used:
1)\[{\text{B = m}}{{\text{c}}^{\text{2}}}\] where ${\text{B}}$ is the binding energy, ${\text{c}}$ is the speed of light and
\[\Delta {\text{m}}\] is the mass defect ---(i)
2) \[{{\Delta m = [N}}{{\text{M}}_{\text{n}}}{\text{ + Z}}{{\text{M}}_{\text{p}}}{\text{ - M}}\left( {{\text{N,Z}}} \right){\text{]}}\] where ${\text{N}}$ is the no. of neutrons,${\text{Z}}$ is the atomic no. Or no. of protons,
\[{{\text{M}}_{\text{n}}}\]is the mass of a neutron, ${{\text{M}}_{\text{p}}}$ is the mass of a proton and ${\text{M}}$ is the practical mass ---(ii)
3) ${\text{N}} = {\text{A}} - {\text{Z}}$ where ${\text{N}}$ is the no. of neutrons, ${\text{A}}$ is the mass number and, ${\text{Z}}$ is the atomic no.
Complete step by step answer:
From the question, we know that we need to calculate the mass defect in terms of the binding energy of an atom.
For this, we need to put the value of equation (ii) in equation (i)
We get,
\[{\text{B = }}\left[ {{\text{N}}{{\text{M}}_{\text{n}}}{\text{ + Z}}{{\text{M}}_{\text{p}}}{\text{ - M}}\left( {{\text{N,Z}}} \right)} \right]{{\text{c}}^{\text{2}}}\]
Now arranging this equation using ${\text{N}} = {\text{A}} - {\text{Z}}$ we now have
\[\dfrac{{\text{B}}}{{{{\text{c}}^{\text{2}}}}}{\text{ = }}\left( {{\text{A - Z}}} \right){{\text{M}}_{\text{n}}}{\text{ + Z}}{{\text{M}}_{\text{p}}} - {\text{M}}\left( {{\text{N,Z}}} \right)\]
Rearranging this further by grouping the common terms
\[ \Rightarrow {\text{M}}\left( {{\text{N,Z}}} \right){\text{ = Z}}{{\text{M}}_{\text{p}}}{\text{ + }}\left( {{\text{A - Z}}} \right){{\text{M}}_{\text{n}}}{\text{ - }}\dfrac{{\text{B}}}{{{{\text{c}}^{\text{2}}}}}\]
We reach the desired equation
\[ \Rightarrow {\text{M}}\left( {{\text{N,Z}}} \right){\text{ = Z}}{{\text{M}}_{\text{p}}}{\text{ + N}}{{\text{M}}_{\text{n}}}{\text{ - }}\dfrac{{\text{B}}}{{{{\text{c}}^{\text{2}}}}}\]
Therefore, we can conclude that the correct answer to this question is option B.
Additional information:
In case of nuclear reactions (splitting or fusion), a small amount of mass of the nucleus gets converted into huge amounts of energy. This implies that this amount of mass is removed from the total mass of the original atom. This energy is distinct from the energy generated by other atomic phenomena. These nuclear binding energies are enormous, million times greater than the electron binding energies of atoms and may prove to be a sustainable energy source in the future.
Note:
Do not confuse between the terms of nuclear mass and chemical mass. Nuclear mass refers to a physical property while chemical mass is based on a convention that carbon has a mass defect of zero. To eliminate this confusion between different disciplines of science chemical mass is now referred to as mass excess.
Formula used:
1)\[{\text{B = m}}{{\text{c}}^{\text{2}}}\] where ${\text{B}}$ is the binding energy, ${\text{c}}$ is the speed of light and
\[\Delta {\text{m}}\] is the mass defect ---(i)
2) \[{{\Delta m = [N}}{{\text{M}}_{\text{n}}}{\text{ + Z}}{{\text{M}}_{\text{p}}}{\text{ - M}}\left( {{\text{N,Z}}} \right){\text{]}}\] where ${\text{N}}$ is the no. of neutrons,${\text{Z}}$ is the atomic no. Or no. of protons,
\[{{\text{M}}_{\text{n}}}\]is the mass of a neutron, ${{\text{M}}_{\text{p}}}$ is the mass of a proton and ${\text{M}}$ is the practical mass ---(ii)
3) ${\text{N}} = {\text{A}} - {\text{Z}}$ where ${\text{N}}$ is the no. of neutrons, ${\text{A}}$ is the mass number and, ${\text{Z}}$ is the atomic no.
Complete step by step answer:
From the question, we know that we need to calculate the mass defect in terms of the binding energy of an atom.
For this, we need to put the value of equation (ii) in equation (i)
We get,
\[{\text{B = }}\left[ {{\text{N}}{{\text{M}}_{\text{n}}}{\text{ + Z}}{{\text{M}}_{\text{p}}}{\text{ - M}}\left( {{\text{N,Z}}} \right)} \right]{{\text{c}}^{\text{2}}}\]
Now arranging this equation using ${\text{N}} = {\text{A}} - {\text{Z}}$ we now have
\[\dfrac{{\text{B}}}{{{{\text{c}}^{\text{2}}}}}{\text{ = }}\left( {{\text{A - Z}}} \right){{\text{M}}_{\text{n}}}{\text{ + Z}}{{\text{M}}_{\text{p}}} - {\text{M}}\left( {{\text{N,Z}}} \right)\]
Rearranging this further by grouping the common terms
\[ \Rightarrow {\text{M}}\left( {{\text{N,Z}}} \right){\text{ = Z}}{{\text{M}}_{\text{p}}}{\text{ + }}\left( {{\text{A - Z}}} \right){{\text{M}}_{\text{n}}}{\text{ - }}\dfrac{{\text{B}}}{{{{\text{c}}^{\text{2}}}}}\]
We reach the desired equation
\[ \Rightarrow {\text{M}}\left( {{\text{N,Z}}} \right){\text{ = Z}}{{\text{M}}_{\text{p}}}{\text{ + N}}{{\text{M}}_{\text{n}}}{\text{ - }}\dfrac{{\text{B}}}{{{{\text{c}}^{\text{2}}}}}\]
Therefore, we can conclude that the correct answer to this question is option B.
Additional information:
In case of nuclear reactions (splitting or fusion), a small amount of mass of the nucleus gets converted into huge amounts of energy. This implies that this amount of mass is removed from the total mass of the original atom. This energy is distinct from the energy generated by other atomic phenomena. These nuclear binding energies are enormous, million times greater than the electron binding energies of atoms and may prove to be a sustainable energy source in the future.
Note:
Do not confuse between the terms of nuclear mass and chemical mass. Nuclear mass refers to a physical property while chemical mass is based on a convention that carbon has a mass defect of zero. To eliminate this confusion between different disciplines of science chemical mass is now referred to as mass excess.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

10 examples of friction in our daily life

