January 9, 2024

Explore comprehensive NCERT Solutions for Class 12 Science Physics Chapter 5 – Nuclei, featuring clear and concise step-by-step explanations. These solutions have gained immense popularity among Class 12 Science students, proving invaluable for completing homework assignments efficiently and preparing for exams. Free access to all questions and answers from the NCERT Book of Class 12 Science Physics Chapter 5 is available here, ensuring a convenient and effective study resource for students. Page No 462: Question 13.1: (a) Two stable isotopes of lithium  and have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium. (b) Boron has two stable isotopes, and. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of  and . ANSWER: (a) Mass of lithium isotope , m1 = 6.01512 u Mass of lithium isotope , m2 = 7.01600 u Abundance of , η1= 7.5% Abundance of , η2= 92.5% The atomic mass of lithium atom is given as: (b) Mass of boron isotope , m1 = 10.01294 u Mass of boron isotope , m2 = 11.00931 u Abundance of , η1 = x% Abundance of , η2= (100 − x)% Atomic mass of boron, m = 10.811 u The atomic mass of boron atom is given as: And 100 − x = 80.11% Hence, the abundance of  is 19.89% and that of is 80.11%. Page No 462: Question 13.2: The three stable isotopes of neon: and have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon. ANSWER: Atomic mass of , m1= 19.99 u Abundance of , η1 = 90.51% Atomic mass of , m2 = 20.99 u Abundance of , η2 = 0.27% Atomic mass of , m3 = 21.99 u Abundance of , η3 = 9.22% The average atomic mass of neon is given as: Page No 462: Question 13.3: Obtain the binding energy (in MeV) of a nitrogen nucleus, given =14.00307 u ANSWER: Atomic mass of nitrogen, m = 14.00307 u A nucleus of nitrogen  contains 7 protons and 7 neutrons. Hence, the mass defect of this nucleus, Δm = 7mH + 7mn − m Where, Mass of a proton, mH = 1.007825 u Mass of a neutron, mn= 1.008665 u ∴Δm = 7 × 1.007825 + 7 × 1.008665 − 14.00307 = 7.054775 + 7.06055 − 14.00307 = 0.11236 u But 1 u = 931.5 MeV/c2 ∴Δm = 0.11236 × 931.5 MeV/c2 Hence, the binding energy of the nucleus is given as: Eb = Δmc2 Where, c = Speed of light ∴Eb = 0.11236 × 931.5  = 104.66334 MeV Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV. Page No 462: Question 13.4: Obtain the binding energy of the nuclei  and in units of MeV from the following data: = 55.934939 u = 208.980388 u ANSWER: Atomic mass of, m1 = 55.934939 u nucleus has 26 protons and (56 − 26) = 30 neutrons Hence, the mass defect of the nucleus, Δm = 26 × mH + 30 × mn − m1 Where, Mass of a proton, mH = 1.007825 u Mass of a neutron, mn = 1.008665 u ∴Δm = 26 × 1.007825 + 30 × 1.008665 − 55.934939 = 26.20345 + 30.25995 − 55.934939 = 0.528461 u But 1 u = 931.5 MeV/c2 ∴Δm = 0.528461 × 931.5 MeV/c2 The binding energy of this nucleus is given as: Eb1 = Δmc2 Where, c = Speed of light ∴Eb1 = 0.528461 × 931.5  = 492.26 MeV Average binding energy per nucleon  Atomic mass of, m2 = 208.980388 u nucleus has 83 protons and (209 − 83) 126 neutrons. Hence, the mass defect of this nucleus is given as: Δm‘ = 83 × mH + 126 × mn − m2 Where, Mass of a proton, mH = 1.007825 u Mass of a neutron, mn = 1.008665 u ∴Δm‘ = 83 × 1.007825 + 126 × 1.008665 − 208.980388 = 83.649475 + 127.091790 − 208.980388 = 1.760877 u But 1 u = 931.5 MeV/c2 ∴Δm‘ = 1.760877 × 931.5 MeV/c2 Hence, the binding energy of this nucleus is given as: Eb2 = Δm‘c2 = 1.760877 × 931.5 = 1640.26 MeV Average bindingenergy per nucleon =  Page No 462: Question 13.5: A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of atoms (of mass 62.92960 u). ANSWER: Mass of a copper coin, m’ = 3 g Atomic mass of atom, m = 62.92960 u The total number of atoms in the coin Where, NA = Avogadro’s number = 6.023 × 1023atoms /g Mass number = 63 g nucleus has 29 protons and (63 − 29) 34 neutrons ∴Mass defect of this nucleus, Δm‘ = 29 × mH + 34 × mn − m Where, Mass of a proton, mH = 1.007825 u Mass of a neutron, mn = 1.008665 u ∴Δm‘ = 29 × 1.007825 + 34 × 1.008665 − 62.9296 = 0.591935 u Mass defect of all the atoms present in the coin, Δm = 0.591935 × 2.868 × 1022 = 1.69766958 × 1022 u But 1 u = 931.5 MeV/c2 ∴Δm = 1.69766958 × 1022 × 931.5 MeV/c2 Hence, the binding energy of the nuclei of the coin is given as: Eb= Δmc2 = 1.69766958 × 1022 × 931.5  = 1.581 × 1025 MeV But 1 MeV = 1.6 × 10−13 J Eb = 1.581 × 1025 × 1.6 × 10−13 = 2.5296 × 1012 J This much energy is required to separate all the neutrons and protons from the given coin. Page No 462: Question 13.6: Write nuclear reaction equations for (i) α-decay of (ii) α-decay of  (iii) β−-decay of (iv) β−-decay of  (v) β+-decay of (vi) β+-decay of  (vii) Electron capture of  ANSWER: α is a nucleus of helium and β is an electron (e− for β− and e+ for β+). In every α-decay, there is a loss of 2 protons and 4 neutrons. In every β+-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β−-decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus. For the given cases, the various nuclear reactions can be written as: Page No 462: Question 13.7: A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value? ANSWER: Half-life of the radioactive isotope = T years Original amount of the radioactive isotope = N0 (a) After decay, the amount of the radioactive isotope = N It is given that only 3.125% of N0 remains after decay. Hence, we can write: Where, λ = Decay constant t = Time Hence, the isotope will take about 5T years to reduce to 3.125% of its original value. (b) After decay, the amount of the radioactive isotope = N It …

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