Chapter 1 - Atomic Structure and Nuclear Physics PDF

MOR|AAS|RIM Chapter 1: Atomic Structure and Nuclear Physics CONCEPT: 1.1 Evolution of Atomic Models Dalton’s Atomic Theory According to the postulates proposed by Dalton, the atomic structure comprises atoms, the smallest particle responsible for the...

MOR|AAS|RIM Chapter 1: Atomic Structure and Nuclear Physics CONCEPT: 1.1 Evolution of Atomic Models Dalton’s Atomic Theory According to the postulates proposed by Dalton, the atomic structure comprises atoms, the smallest particle responsible for the chemical reactions to occur. Postulates of Dalton’s Atomic Theory The following are the postulates of his theory: Every matter is made up of atoms. Atoms are indivisible. Specific elements have only one type of atom in them. Each atom has its own constant mass that varies from element to element. Atoms can neither be created nor be destroyed but can be transformed from one form to another. Dalton’s atomic theory successfully explained the Laws of chemical reactions, namely, the Law of conservation of mass, Law of constant properties, Law of multiple proportions, and Law of reciprocal proportions. Discovery of subatomic particles electrons, protons, and neutrons discarded the indivisible nature of the atom proposed by John Dalton. Thomson's Atomic model Thomson atomic model was proposed by William Thomson in the year 1900. This model explained the description of an inner structure of the atom theoretically. It was strongly supported by Sir Joseph Thomson, who had discovered the electron earlier. Thomson assumed that an electron is two thousand times lighter than a proton and believed that an atom is made up of thousands of electrons. In this atomic structure model, he considered atoms surrounded by a cloud having positive as well as negative charges. The demonstration of the ionization of air by X-ray was also done by him together with Rutherford. They were the first to demonstrate it. Thomson’s model of an atom is similar to a plum pudding. 1 Chapter 1: Atomic Structure and Nuclear Physics Postulates of Thomson’s atomic model Postulate 1: An atom consists of a positively charged sphere with electrons embedded in it Postulate 2: An atom as a whole is electrically neutral because the negative and positive charges are equal in magnitude Limitations of Thomson’s atomic model It failed to explain the stability of an atom because his model of atom failed to explain how a positive charge holds the negatively charged electrons in an atom. Therefore, This theory also failed to account for the position of the nucleus in an atom Thomson’s model failed to explain the scattering of alpha particles by thin metal foils No experimental evidence in its support Although Thomson’s model was not an accurate model to account for the atomic structure, it proved to be the base for the development of other atomic models. The study of the atom and its structure has paved the way for numerous inventions that have played a significant role in the development of humankind. Rutherford Atomic Model and Limitations Rutherford Atomic Model – The plum pudding model given by J. J. Thomson failed to explain certain experimental results associated with the atomic structure of elements. Ernest Rutherford, a British scientist conducted an experiment and based on the observations of this experiment, he explained the atomic structure of elements and proposed Rutherford’s Atomic Model. 2  MOR|AAS|RIM Rutherford’s Alpha Scattering Experiment Rutherford conducted an experiment by bombarding a thin sheet of gold with α-particles and then studied the trajectory of these particles after their interaction with the gold foil. Rutherford, in his experiment, directed high energy streams of α-particles from a radioactive source at a thin sheet (100 nm thickness) of gold. In order to study the deflection caused to the α-particles, he placed a fluorescent zinc sulphide screen around the thin gold foil. Rutherford made certain observations that contradicted Thomson’s atomic model. Observations of Rutherford’s Alpha Scattering Experiment The observations made by Rutherford led him to conclude that: 1. A major fraction of the α-particles bombarded towards the gold sheet passed through the sheet without any deflection, and hence most of the space in an atom is empty. 2. Some of the α-particles were deflected by the gold sheet by very small angles, and hence the positive charge in an atom is not uniformly distributed. The positive charge in an atom is concentrated in a very small volume. 3. Very few of the α-particles were deflected back, that is only a few α-particles had nearly 180o angle of deflection. So the volume occupied by the positively charged particles in an atom is very small as compared to the total volume of an atom. 3 Chapter 1: Atomic Structure and Nuclear Physics Based on the above observations and conclusions, Rutherford proposed the atomic structure of elements. According to the Rutherford atomic model: 1. The positive charge and most of the mass of an atom is concentrated in an extremely small volume. He called this region of the atom as a nucleus. 2. Rutherford’s model proposed that the negatively charged electrons surround the nucleus of an atom. He also claimed that the electrons surrounding the nucleus revolve around it with very high speed in circular paths. He named these circular paths as orbits. 3. Electrons being negatively charged and nucleus being a densely concentrated mass of positively charged particles are held together by a strong electrostatic force of attraction. Limitations of Rutherford Atomic Model Although the Rutherford atomic model was based on experimental observations, it failed to explain certain things. Rutherford proposed that the electrons revolve around the nucleus in fixed paths called orbits. According to Maxwell, accelerated charged particles emit electromagnetic radiations and hence an electron revolving around the nucleus should emit electromagnetic radiation. This radiation would carry energy from the motion of the electron which would come at the cost of shrinking of orbits. Ultimately the electrons would collapse in the nucleus. Calculations have shown that as per the Rutherford model, an electron would collapse into the nucleus in less than 10-8 seconds. So the Rutherford model was not in accordance with Maxwell’s theory and could not explain the stability of an atom. One of the drawbacks of the Rutherford model was also that he did not say anything about the arrangement of electrons in an atom which made his theory incomplete. Although the early atomic models were inaccurate and failed to explain certain experimental results, they formed the base for future developments in the world of quantum mechanics. Niels Bohr Atomic Model And Limitations Thomson’s atomic model and Rutherford’s atomic model failed to answer any questions related to the energy of an atom and its stability. In the year 1913, Niels Bohr proposed an atomic structure model, describing an atom as a small, positively charged nucleus surrounded by electrons that travel in circular orbits around the positively charged nucleus like planets around the sun in our solar system, with attraction provided by electrostatic forces, popularly known as Bohr’s atomic model. It was basically an improved version of Rutherford’s atomic model overcoming its limitations. On most of the points, he is in agreement with him, like concepts of nucleus and electrons orbiting it. Salient features of Niels Bohr atomic model are: 4  MOR|AAS|RIM Electrons revolve around the nucleus in stable orbits without emission of radiant energy. Each orbit has a definite energy and is called an energy shell or energy level. An orbit or energy level is designated as K, L, M, N shells. When the electron is in the lowest energy level, it is said to be in the ground state. An electron emits or absorbs energy when it jumps from one orbit or energy level to another. When it jumps from a higher energy level to lower energy level it emits energy while it absorbs energy when it jumps from a lower energy level to a higher energy level. The energy absorbed or emitted is equal to the difference between the energies of the two energy levels (E1, E2) and is determined by Plank’s equation. ΔE = E2-E1 = h𝜈 Where, ΔE = energy absorbed or emitted h= Plank’s constant 𝜈= frequency of electromagnetic radiation emitted or absorbed The angular momentum of an electron revolving in energy shells is given by: mevr = nh/2π Where, n= number of corresponding energy shell; 1, 2, 3 ….. me= mass of the electron v= velocity r=radius h= Plank’s constant 5 Chapter 1: Atomic Structure and Nuclear Physics Limitations of Bohr Atomic Model Theory It violates the Heisenberg Uncertainty Principle. The Bohr atomic model theory considers electrons to have both a known radius and orbit i.e. known position and momentum at the same time, which is impossible according to Heisenberg. The Bohr atomic model theory made correct predictions for smaller sized atoms like hydrogen, but poor spectral predictions are obtained when larger atoms are considered. It failed to explain the Zeeman effect when the spectral line is split into several components in the presence of a magnetic field. It failed to explain the Stark effect when the spectral line gets split up into fine lines in the presence of an electric field. CONCEPT: 1.2 Further Developments in Atomic Models QUANTUM MECHANICAL MODEL The quantum mechanical model of the atom is a fundamental theory in physics that describes the behavior of electrons in atoms using principles of quantum mechanics. The quantum mechanical model represents a profound shift from classical physics and provides a more accurate and comprehensive understanding of atomic structure and behavior. It lays the foundation for modern chemistry and physics, influencing various fields including quantum chemistry, materials science, and nanotechnology. Modern Atomic Theory Modern atomic theory builds on earlier models, integrating principles from quantum mechanics to provide a comprehensive understanding of atomic structure and behavior. Modern atomic theory has transformed our understanding of matter at the atomic and subatomic levels, bridging chemistry, physics, and materials science. It provides a robust framework for exploring and explaining the behavior of atoms and molecules in various contexts, from the microscopic to the macroscopic scale. CONCEPT: 1.3 Atomic Structure Nucleus of an Atom The nucleus of an atom is the central region of an atom where the majority of the mass is concentrated. Through the scattering of alpha particles experiment by Rutherford, we learned that the nucleus of an atom contains a majority of the mass of the atom. Numerically speaking, the nucleus of an atom occupies almost 10-14 times the volume of the atom but contains 99.99% of the atomic mass. The 6  MOR|AAS|RIM nucleus of an atom is so small that if you expanded an atom to fill up a room, the nucleus of an atom would still be no larger than a pinhead! Symbol of Nucleus Number of protons inside the nucleus is called atomic number (Z) and total number of protons and neutrons is called mass number (A). Subtraction of A and Z is the number of neutron (N). That means, N = A – Z. A nucleus is normally represented by 𝐴𝑍𝑋 Atomic Number (Z) The number of protons present in the nucleus of an atom is equal to its atomic number (Z). Thus Atomic number (Z) = number of protons in the nucleus of an atom Example Atomic number of Carbon(C) = 6 = number of protons 7 Chapter 1: Atomic Structure and Nuclear Physics Atomic Mass (A) Atomic mass is equal to the sum of protons and neutrons present in the nucleus of the atom. Thus Atomic mass = number of protons + number of neutrons For Example Atomic mass of carbon = 12 Different Atomic Species Isotopes Atoms of the same element which have the same atomic number (or number of protons) but have different mass numbers (or number of neutrons) are called isotopes. Isotones Isotones are two or more species of atoms or nuclei that have the same number of neutrons. Their atomic number(Z) and mass number (A) are different. But the value A-Z( number of neutrons) is the same. Example 𝟑𝟕 𝟑𝟗 𝟏𝟕𝑪𝒍 and 𝟏𝟗𝑲 are isotones. The atomic nucleus of Chlorine consists of 17 protons and 20 neutrons. Whereas the nucleus of the species Potassium contains 19 protons and 20 neutrons. 8  MOR|AAS|RIM Isobars Atoms of different elements that have the same mass number but different atomic numbers are called Isobars. CONCEPT: 1.4 Nuclear Force and Nuclear Energy Nuclear Force The nuclear force is one of the four fundamental forces of nature, the others being gravitational and electromagnetic forces. In fact, being 10 million times stronger than the chemical binding forces, they are also known as the strong forces. In this section, we will discuss this force in detail. We can define nuclear force as: The nuclear force is a force that acts between the protons and neutrons of atoms. The nuclear force is the force that binds the protons and neutrons in a nucleus together. This force can exist between protons and protons, neutrons and protons or neutrons and neutrons. This force is what holds the nucleus together. The charge of protons, which is +1e, tends to push them away from each other with a strong electric field repulsive force, following Coulomb’s law. But nuclear force is strong enough to keep them together and to overcome that resistance at short range. 9 Chapter 1: Atomic Structure and Nuclear Physics Properties of Nuclear Force It is attractive in nature but with a repulsive core. That is the reason that the nucleus is held together without collapsing in itself. The range of a nuclear force is very short. At 1 Fermi, the distance between particles in a nucleus is tiny. At this range, the nuclear force is much stronger than the repulsive Coulomb’s force that pushes the particles away. However, if the distance is anything more than 2.5 Fermi, nuclear force is practically non-existent. The nuclear force is identical for all nucleons. It does not matter if it is a neutron or proton, once the Coulomb resistance is taken into consideration, nuclear force affects everything in the same way. At a distance of less than 0.7 Fermi, this force becomes repulsive. It is one of the most interesting properties of nuclear force, as this repulsive component of the force is what decides the size of the nucleus. The nucleons come closer to each other to the point that the force allows, after which they cannot come any closer because of the repulsive property of the force. Nuclear Force Examples The most obvious example of Nuclear Force, as discussed earlier, is the binding of protons, which are repulsive in nature because of their positive charge. Nuclear Energy Nuclear Energy is the energy in the core of an atom. Where an atom is a tiny particle that constitutes every matter in the universe. Normally, the mass of an atom is concentrated at the centre of the nucleus. Neutrons and Protons are the two subatomic particles that comprehend the nucleus. There is an exact massive amount of energy in bonds that bind atoms together. Nuclear Energy is discharged by nuclear reactions either by fission or fusion. In nuclear fusion, atoms combine to form a larger atom. In nuclear fission, the division of atoms takes place to form smaller atoms by releasing energy. Nuclear power plants produce energy using nuclear fission. The Sun produces energy using the mechanism of nuclear fusion. Applications of Nuclear Energy Nuclear technology Nuclear medicine Nuclear Technology is used in Industries Agricultural uses of nuclear technology 10  MOR|AAS|RIM Environmental uses of nuclear technology Biological Experimentations Medical diagnosis and treatments Scientific Investigations Engineering Projects Neutron Activation Analysis CONCEPT: 1.5 Mass Defect and Binding Energy Mass Defect Mass defect is the difference between the actual atomic mass and the predicted mass calculated by adding the mass of protons and neutrons present in the nucleus. The actual atomic mass is less than the predicted mass calculated by adding the masses of nucleons. This additional mass is accounted for by binding energy that is released when a nucleus is formed. When a nucleus is formed, some of the mass is converted to energy and this results in the mass defect. Due to this reason, the actual mass of an atomic nucleus is less than the mass of particles it is made up of. The actual mass of the atomic nucleus is always less than the mass of protons and neutrons present in the nucleus. When a nucleus is formed, energy is released. This energy is removed in the form of a reduction in total mass. This missing mass is known as the ‘mass defect’ and it accounts for the energy released. The mass defect (𝚫M) can be calculated by subtracting the original atomic mass (MA) from the sum of the mass of protons (mp= 1.00728 amu) and neutrons (mn= 1.00867 amu) present in the nucleus. Mass defect formula: 𝚫M = (Zmp + Nmn) – MA 𝚫M – mass defect MA – mass of the nucleus mp – mass of a proton, i.e. 1.00728 amu mn – mass of a neutron, i.e. 1.00867 amu Z – number of protons N – number of neutrons Binding Energy Binding energy is typically defined as the smallest amount of energy that is required to remove a particle from a system of particles. In other words, it is the energy that is used to separate a system of particles into single units. We study about binding energy mostly in atomic physics and chemistry as well as in condensed matter physics. In nuclear physics, the term binding energy is used to describe separation energy. Binding energy is necessary to split subatomic particles in atomic nuclei or the nucleus of an atom into its components, namely: neutrons and protons or collectively known as nucleons. The binding energy of nuclei is a positive value because every nucleus needs net energy to isolate them into neutron and proton. Binding energy is also applicable to atoms and ions bound together in crystals. Applications 11 Chapter 1: Atomic Structure and Nuclear Physics Binding energy is also applied in determining whether fusion or fission will be favourable. For elements that are lighter than iron-56, the fusion releases energy since the nuclear binding energy rises with the hike in mass. Elements that are heavier than iron-56 release energy on fission since the lighter elements consist of higher binding energy. Hence, there exists a peak at iron-56 according to the nuclear binding energy curve. Binding Energy Formula Energy equivalent to mass defect is termed as binding energy. A nucleus is like an inflexible spherical ball moulded by getting together a huge amount of miniature spherical balls in the nucleons form. Something is required to bound the nucleons collected. It is that something which assists as the glue. Every nucleon has to donate some of its mass, to deliver the energy resulting in a mass defect. Binding Energy is also defined as the energy required to break down a nucleus into its component nucleons. Binding Energy Formula Binding Energy = mass defect x c2 Where, C = Speed of light in vacuum = 2.9979 x 108 m/s. Binding Energy is expressed in terms of kJ/mole of nuclei or MeV’s/nucleon. Binding Energy Calculation 1) Calculate the Binding Energy of the Deuteron. Give data: a mass of the deuteron is 1875.61MeV/c2 or 3.34359✕10-27 Kg. Solution: Given: Mass of deuteron mD = 1875.61MeV/c2 or 3.34359✕10-27 Kg. Atomic Mass number of Deuteron; A=2 The atomic number of Deuteron; Z=1 Mass defect Δm =? Binding energy Eb = ? 12  MOR|AAS|RIM Formula used: Mass defect is given by Δm=Zmp+(A−Z)mn−mnuc Binding energy is given by Eb=(Δm)c2 Calculation: Mass defect is given by ΔM=ZMp+(A−Z)Mn−Mnuc Here Z=1 and (A-Z)=1, Mn =939.57MeV/c2 and Mp =938.28 MeV/c2. Substituting the values we get- ΔM = Mp + Mn– MD = 938.28 MeV/c2 + 939.57MeV/c2 – 1875.61MeV/c2 =2.24 MeV/c2 Thus, the mass defect is 2.24 MeV/c2 The binding energy of the Deuteron is thus given by Eb=(Δm)c2 = (2.24 MeV/c2)(c2) =2.24 MeV A minimum of 2.24 million electron volt energy is required to break Deuteron into Proton and Neutron. This is a very large value. However, the energy required to separate an electron from a hydrogen atom by overcoming electromagnetic force(Coulomb force) is approximately 10 eV. This comparison clearly indicates the strength of the nuclear force. Although its range lies in multiple Femto meters, It is one of the strongest forces in nature. Binding Energy per Nucleon The total energy required by the nucleus to release its nucleon is called binding energy per nucleon. It is the foundation of nuclear power technology. Solved Examples Problem 1: Calculate the binding energy per nucleon for an alpha particle whose mass defect is calculated as 0.0292amu. Given: mass defect = 0.0292amu Convert the mass defect into kg (1 amu = 1.6606 x 10-27 kg) Mass defect =(0.0292 )( 1.6606 x 10-27 )= 0.04848 x 10-27 kg/nucleus Convert this mass into energy using DE = Dmc2, where c = 2.9979 x 108 m/s. E = (0.04848 x 10-27)(2.9979 x 108 )2 = 0.4357 x 10-11 J/nucleus Convert Energy in terms of kJ/mole (1 kJ = 1000 J) convert to mole by multiplying with the Avogadro number (6.022 x 1023 nuclei/mol) Therefore, E = (0.4357 x 10-11)(6.022 x 1023)/1000 Binding Energy E = 2.62378 x 109 kJ/mole 13 Chapter 1: Atomic Structure and Nuclear Physics Mathematical Problems – 1.6 1) The electron in a hydrogen atom is rotating in the third orbit. What is its angular momentum? (h=6.6×10⁻³⁴ Js) Solution: According to Bohr's postulate, angular momentum of an electron is: L = nħ/2π = (3 × 6.6 × 10⁻³⁴) / (2 × 3.14) = 3.15 × 10⁻³⁴ Js 2) Find (i) the mass defect, (ii) the binding energy and (iii) the binding energy per nucleon for a helium (He) nucleus. Express mass defect in amu unit and binding energy in MeV, eV and Joule units. [Mass of a proton = 1.00728 a.m.u., mass of a neutron = 1.00876 a.m.u. and 1 a.m.u. = 931 MeV]. Solution: We know, there are 2 protons and 2 neutrons in a helium nucleus. Now, mass of 2 protons = 2 × 1.00728 = 2.014556 a.m.u. Mass of 2 neutrons = 2 × 1.00876 = 2.01734 a.m.u. Their total mass = 2.014556 + 2.01734 = 4.03190 a.m.u. But actual mass of a helium nucleus = 4.00276 a.m.u. (i) Mass defect. Δm = 4.03190 - 4.00276 = 0.02914 a.m.u. (ii) So, binding energy = Δm × 931 = 0.02914 × 931 = 27.129 MeV = 27.129 × 10⁶ eV = 27.129 × 10⁶ × 1.6 × 10⁻¹⁹ J = 4.34 × 10⁻¹² J (iii) Number of nucleon in helium nucleus A = 4 So, binding energy per nucleon = (B.E. / A) = (27127 / 4) = 6782 MeV = 6.782 x 10⁶ eV = 6.782 x 10⁶ x 1.6 x 10⁻¹⁹ J = 1.085 x 10⁻¹² J 3) Calculate mass defect and binding energy of Li⁷ nucleus. Given, mₙ = 1.008665 amu, mₚ = 1.007277 amu, mass of Lithium nucleus = 7.016005 amu and 1 amu = 1.66 x 10⁻²⁷ kg. Solution: We know: Δm = Zmₚ + (A - Z)mₙ - M or, Δm = (3 x 1.007277) + (7 - 3) x (1.008665) - 7.016005 = (3.021831 + 4.034660) - 7.016005 = 7.048491 - 7.016006 = 0.032486 amu = 0.032486 x (1.66 x 10⁻²⁷ kg) = 0.044 x 10⁻²⁷ kg Again, Binding energy, E = Δmc² = 0.044 x 10⁻²⁷ kg x (3 x 10⁸ m/s)² = 3.96 x 10⁻¹¹ J 14  MOR|AAS|RIM Chapter 1: Practice Problems 1. What will be the frequency of the photon emitted when a Hydrogen atom comes to ground state from the excited state? The energies at excited the first Bohr's ground states are respectively -3.4 eV and -13.6 eV. Ans: 2.46 x 1015 Hz 2. The symbol of aluminum nucleus is 27 13𝐴𝑙. What are the number of protons, neutron, mass number and atomic number of this nucleus? Ans: 13,14,27 and 13 3. Find the energy of an electron in the first Bohr's orbit in a hydrogen atom. Given that, mass and charge of electron are respectively 9.1 × 10 - 31 kg and 16 × 10 - 19 C. Permittivity of vacuum epsilon εo = 8.85 × 10 - 12 C 2 N - 1 m - 2. Ans: -13.6eV 4. Find the quantum number of Bohr's orbit of a hydrogen atom whose radius is 0.01 mm. What is the energy of a hydrogen atom at this state? The radius of the first Bohr's orbit is 0.53 Å and energy of the ground state is -13.6 eV. Ans: 434; -7.22 x 10-5eV 5. Find the radius of third Bohr's orbit in a hydrogen atom. Calculate the energy of the electron of that orbit. Ans: 4.786Å; -1.5eV 6. The energy of the ground state of hydrogen atoms - 13.6eV Find the energy of the photon emitted in the transition from n = 4 to n = 2. Ans: 2.55 eV 7. The Hα line of Balmer series is obtained from the transition n=3 (energy= - 1.5eV) to n=2 (energy= - 3.4eV). Calculate the wavelength for line. Ans: 6513 Å 8. The first line of the Lyman series in the hydrogen spectrum has a wavelength 1200 A. Calculate the wavelength of the second line. Ans: 1012.5 Å 9. Find the frequency of the photon emitted by the hydrogen atom when it comes from-15 eV energy state to-3.4 eV state? Ans: 4.59 x 1014 Hz 10. Find the radius of the first orbit of hydrogen atom. Ans: 0.53 Å 11. Find the radius of second Bohr's orbit in hydrogen atom. Calculate the energy of the electron of that orbit. Ans: 2.127 Å; -3.383eV 12. Find the mass defect, binding energy and binding energy per nucleon of a helium nucleus. Given that, mass of neutron 1.008665 u, atomic mass of hydrogen = 1.007825 u and atomic mass of helium = 4.002603u. Ans: 0.030377amu, 28.296MeV,7.07MeV per nucleon 13. Find the binding energy and binding energy per nucleon MeV, eV and joule unit for 56 26𝐹𝑒 nucleus. Here the atomic mass of hydrogen, mH = 1007825 u, mass of a neutron, mn = 1.008665, atomic mass of iron, M atom =55.834942u. Ans:492.26MeV, 8.79MeV per nucleon 14. An information table has been given below: Element Number of Mass number Mass of nucleus 1amu= 931MeV protons amu U 92 235 235.0439 mp =1.00728amu C 6 12 12.0000 mn =1.00876amu c = 3× 108 ms-1 Fe 26 56 56.0000 He 2 4 4.00276 (a) Find the mass defect of uranium. (b) Using the information given in the stem draw graph of binding energy per nucleon verse the mass number. Ans: (a) 1.87854amu (b) U [7.4422 MeV per nucleon], C[ 7.4462 MeV per nucleon], Fe[ 7.5153 MeV per nucleon], He[ 7.4666 MeV per nucleon] 15. Find the binding energy and binding energy per nucleon of 238 92𝑈 nucleus in MeV and joule unit. Given that, atomic mass of hydrogen = 1.007825u, mass of a neutron = 1.008665 atomic mass of neutron = 238.05078u. Ans: 1802MeV, 7.51 MeV per nucleon. 15 Chapter 1: Atomic Structure and Nuclear Physics 16. Find the mass defect, binding energy and binding energy per nucleon of 6228𝑁𝑖 nucleus Given that, atomic mass of hydrogen = 1.007825u. Mass of a neutron = 1.008665 u and atomic mass of nickel =61.928348u. Ans: 0.58536u, 545.26MeV, 8.79MeV per nucleon 17. Calculate the binding energy per nucleon for 20 56 238 10𝑁𝑒 , 26Fe and 92𝑈. Given that mass of neutron is 1.008665 amu, mass of proton is 1.007825 amu, mass of 20 56 10𝑁𝑒 is 19.9924 amu, mass of 26Fe is 238 55.93492 amu, 92𝑈 is 238.050783 amu. Ans: Ne = 8.03MeV, Fe = 8.76MeV, U = 7.57MeV 18. The binding energy of 3517𝐶𝑙 nucleus is 298 MeV. Find its atomic mass. Given, mass of a proton (mp)=1.007825amu, mass of neutron (mn)=1.008665amu. Ans: 34.9597amu 19. The mass of 42𝐻𝑒 nucleus is 0.03 amu less than the sum of the masses of 2 protons and 2 neutrons. What is the binding energy per nucleon? Ans: 27.945MeV, 6.986MeV 20. Find Binding energy of an α−particle in MeV? [mproton=1.007825 amu, mneutron=1.008665 amu, mhelium=4.002800 amu] Ans: 28.41 MeV 21. The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu, the binding energy of the helium nucleus will be [1 amu= 931 MeV] Ans: 28.4MeV 22. The binding energy per nucleon of 35 35 17𝐶𝑙 nucleus is( 17𝐶𝑙 =34.98000 amu, mP=1.007825 amu, mn=1.008665 amu and 1 amu is equivalent to 931MeV). Ans: 8.22 MeV 23. The binding energy per nucleon of 4020𝐶𝑎 is ––––––––– ( 40 20𝐶𝑎 =39.962589 amu, m p =1.007825 amu; mn=1.008665amu and 1amu is equivalent to 931.5MeV). Ans: 8.55 MeV 24. Calculate the mass defect and binding energy per nucleon of 168𝑂 which has a mass 15.99491 amu. Mass of neutron =1.008655 amu, Mass of proton =1.007277 amu, Mass of electron =0.0005486 amu, 1amu=931.5 MeV Ans: −0.13254amu, 7.716 Mev/nucleon 25. Atomic mass of 126𝐶 is 13.00335 amu and its mass number is 13.0. If amu =931MeV, binding energy of the neutrons present in the nucleus is? Ans: 3.10MeV 16