January 8, 2024

Explore the comprehensive NCERT Solutions for Class 12 Science Physics Chapter 4 on Atoms, featuring clear, step-by-step explanations. Widely favored by class 12 Science students, these Physics Atoms Solutions serve as a valuable resource for efficiently completing homework assignments and preparing for examinations. You can access all the questions and answers from the NCERT Book of Class 12 Science Physics Chapter 4 at no cost. These solutions are designed to assist you in grasping the concepts quickly and effectively, making your study sessions more productive. Page No 435: Question 12.1: Choose the correct alternative from the clues given at the end of the each statement: (a) The size of the atom in Thomson’s model is ………. the atomic size in Rutherford’s model. (much greater than/no different from/much less than.) (b) In the ground state of ………. electrons are in stable equilibrium, while in ………. electrons always experience a net force. (Thomson’s model/ Rutherford’s model.) (c) A classical atom based on ………. is doomed to collapse. (Thomson’s model/ Rutherford’s model.) (d) An atom has a nearly continuous mass distribution in a ………. but has a highly non-uniform mass distribution in ………. (Thomson’s model/ Rutherford’s model.) (e) The positively charged part of the atom possesses most of the mass in ………. (Rutherford’s model/both the models.) ANSWER: (a) The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude. (b) In the ground state of Thomson’s model, the electrons are in stable equilibrium. However, in Rutherford’s model, the electrons always experience a net force. (c) A classical atom based on Rutherford’s model is doomed to collapse. (d) An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model. (e) The positively charged part of the atom possesses most of the mass in both the models. Page No 435: Question 12.2: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect? ANSWER: In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 × 10−27 kg) is less than the mass of incident α−particles (6.64 × 10−27 kg). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α−particles would not bounce back if solid hydrogen is used in the α-particle scattering experiment. Page No 436: Question 12.3: What is the shortest wavelength present in the Paschen series of spectral lines? ANSWER: Rydberg’s formula is given as: Where, h = Planck’s constant = 6.6 × 10−34 Js c = Speed of light = 3 × 108 m/s (n1 and n2 are integers) The shortest wavelength present in the Paschen series of the spectral lines is given for values n1 = 3 and n2 = ∞. Page No 436: Question 12.4: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level? ANSWER: Separation of two energy levels in an atom, E = 2.3 eV = 2.3 × 1.6 × 10−19 = 3.68 × 10−19 J Let ν be the frequency of radiation emitted when the atom transits from the upper level to the lower level. We have the relation for energy as: E = hv Where, h = Planck’s constant Hence, the frequency of the radiation is 5.6 × 1014 Hz. Page No 436: Question 12.5: The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state? ANSWER: Ground state energy of hydrogen atom, E = − 13.6 eV This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy. Kinetic energy = − E = − (− 13.6) = 13.6 eV Potential energy is equal to the negative of two times of kinetic energy. Potential energy = − 2 × (13.6) = − 27 .2 eV Page No 436: Question 12.6: A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of the photon. ANSWER: For ground level, n1 = 1 Let E1 be the energy of this level. It is known that E1 is related with n1 as: The atom is excited to a higher level, n2 = 4. Let E2 be the energy of this level. The amount of energy absorbed by the photon is given as: E = E2 − E1 For a photon of wavelengthλ, the expression of energy is written as: Where, h = Planck’s constant = 6.6 × 10−34 Js c = Speed of light = 3 × 108 m/s And, frequency of a photon is given by the relation, Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 × 1015 Hz. Page No 436: Question 12.7: (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels. ANSWER: (a) Let ν1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, ν1 is given by the relation, Where, e = 1.6 × 10−19 C ∈0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2 h = Planck’s constant = 6.62 × 10−34 Js For level n2 = 2, we can write the relation for the corresponding orbital speed as: And, for n3 = 3, we can write the relation for the corresponding orbital speed as: Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 × 106 m/s, 1.09 × 106 m/s, 7.27 × 105 m/s respectively. (b) Let T1 be the orbital period of the electron when it is in level n1 = 1. Orbital period is related to orbital speed as: Where, r1 = Radius of the orbit h = Planck’s constant = 6.62 × 10−34 Js e = Charge on an electron = 1.6 × 10−19 C ∈0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2 m = Mass …

NCERT Solutions for Class 12 Science Physics Chapter 4 – Atoms Read More »