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Exercises
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 nonuniform 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.
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.
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 × 10^8 m/s (n 1 and n 2 are integers) The shortest wavelength present in the Paschen series of the spectral lines is given for values n 1 = 3 and n 2 = ∞.
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, n 1 = 1 Let E 1 be the energy of this level. It is known that E 1 is related with n 1 as:
The atom is excited to a higher level, n 2 = 4. Let E 2 be the energy of this level.
The amount of energy absorbed by the photon is given as: E = E 2 − E 1
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 × 10^8 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 × 10^15 Hz.
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 ν 1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n 1 = 1. For charge (e) of an electron, ν 1 is given by the relation,
Where, e = 1.6 × 10−^19 C 0 = Permittivity of free space = 8.85 × 10−^12 N−^1 C^2 m−^2 h = Planck’s constant = 6.62 × 10−^34 Js
For level n 2 = 2, we can write the relation for the corresponding orbital speed as:
For level n 2 = 2, we can write the period as:
Where, r 2 = Radius of the electron in n 2 = 2
And, for level n 3 = 3, we can write the period as:
Where, r 3 = Radius of the electron in n 3 = 3
Hence, the orbital period in each of these levels is 1.52 × 10−^16 s, 1.22 × 10−^15 s, and
4.12 × 10−^15 s respectively.
Question 12.8: The radius of the innermost electron orbit of a hydrogen atom is 5.3 ×10−^11 m. What are the radii of the n = 2 and n =3 orbits?
Answer The radius of the innermost orbit of a hydrogen atom, r 1 = 5.3 × 10−^11 m. Let r 2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as:
For n = 3, we can write the corresponding electron radius as:
Hence, the radii of an electron for n = 2 and n = 3 orbits are 2.12 × 10−^10 m and 4. × 10−^10 m respectively.
Question 12.9: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted? Answer It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV. When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV. Orbital energy is related to orbit level (n) as:
If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:
This radiation corresponds to the Balmer series of the hydrogen spectrum. Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.
Question 12.10: In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 10^11 m with orbital speed 3 × 10^4 m/s. (Mass of earth = 6.0 × 10^24 kg.) Answer Radius of the orbit of the Earth around the Sun, r = 1.5 × 10^11 m Orbital speed of the Earth, ν = 3 × 10^4 m/s Mass of the Earth, m = 6.0 × 10^24 kg According to Bohr’s model, angular momentum is quantized and given as:
Where, h = Planck’s constant = 6.62 × 10−^34 Js n = Quantum number
Hence, the quanta number that characterizes the Earth’ revolution is 2.6 × 10^74.
