Class 12 Physics Chapter 12 Important Questions Atoms

Q 1:- Define impact parameter. Ans:- Impact parameter of an alpha particle is defined as the perpendicular distance of the velocity vector of α-partic

Science is a complex and challenging subject, as it involves so many principles and concepts that are difficult to memorize. Those student who opt for science have to face many challenges and work hard to get good marks in the exam. In this lesson, students will learn about Atoms. The best solution of the problem is to practice as many Physics Class 12 Chapter 12 Important Questions as possible to clear the doubts.

Conceptual Questions  for Class 12 Physics Chapter 12 Atoms

Q 1:- Define impact parameter.
Ans:- Impact parameter of an alpha particle is defined as the perpendicular distance of the velocity vector of α-particle from the central line of gold nucleus, when it is far away from gold nucleus.

Q 2:- What is the scattering angle if impact parameter (b) Has a zero value?
Ans:- If impact parameter b is zero, then the angle of scattering, θ = 180° or π rad.

Q 3:- What force is responsible for an α-particle scattering?
Ans:- The electrostatic repulsive force between nucleus of gold and the alpha particle is responsible for an α-particle scattering.

Q 4:- What is the main feature of Rutherford's nuclear atom model?
Ans:- Rutherford's atom is an electrically neutral sphere consisting of a very small massive and positively charged nucleus at the centre surrounded by revolving electrons in their respective orbits. The requisite centripetal force for orbital motion of electrons is provided by coulombian attractive force between nucleus and the revolving electron.

Q 5:- Explain, how Rutherford's experiment on scattering of α-particles led to the estimation of the size of the nucleus
Ans:- By experimental studies of large angle scattering of α-particles, the value of the distance of the closest approach (r0) for α- particles was determined. The size of nucleus can be either less than or equal to the distance of the closest approach. Thus, distance of the closest approach provides us an upper limit of the size of nucleus. In this way, size of the nucleus was estimated. The estimated size of the nucleus is of the order of 10-14 m.

Q 6:- State Bohr's postulate of quantisation of angular momentum of the election in hydrogen atom.
Ans:- As per the Bohr's postulate, the electron revolves around the nucleus only in those orbits for which angular momentum of electron is some integer multiple of h/2n, where h is the universal Planck's constant.

Mathematically,

`mv_{n}r_{n}=\frac{nh}{2\pi}` where n = 1, 2, 3, ...

Q 7:- What is Bohr's radius? What is its value?
Ans:- Bohr's radius is the size (radius) of the innermost orbit of electron in its ground state (n = 1) for hydrogen atom. Its value is 5.29 x 10-11 m (~5.3 × 10-11 m).

Read also: Atoms Class 12 Physics Notes Chapter 12

Q 8:- Define ionisation energy. What is its value for a hydrogen atom?
Ans:- The minimum amount of energy required to free the valence electron from its ground state in an atom is called 'ionisation energy' for that atom/element. Ionisation energy of hydrogen is +13.6 eV.

Q 9:- How are ionisation energy and ionisation potential interrelated?
Ans:- Numerical value of ionisation energy of an atom expressed electron volts is equal to the ionisation potential for the atom. As an example, ionisation energy for a hydrogen atom is 13.6 eV and its ionising potential is 13.6 V.

Q 10:- What do you mean by Hα, Hβ, Hγ, ... lines of hydrogen spectrum?
Ans:- The first, second, third,... lines of Balmer series of hydrogen spectrum are generally termed as Hα, Hβ, Hγ, ... lines.

Q 11:- What are hydrogenic atoms? Give examples too.
Ans:- Hydrogenic atoms are the atoms consisting of a nucleus with positive charge +Ze and a single electron, where Z = proton number (atomic number) of the nucleus. Hydrogen atom (H), singly ionised helium (He+), doubly ionised lithium (Li++), etc., are the examples of hydrogenic atoms.

Q 12:- How can you say that most of an atom is empty?
Ans:- The size of an atom is of the order of (10-10 - 10-11 m) and the size of a nucleus is of the order of (10-14 - 10-15 m). Outside the nucleus, there are only few electrons orbiting around the nucleus. It conclusively proves that most of an atom is empty. Again, it is due to this reason that most α-particles pass without any deviation in Rutherford's α-particles scattering experiment.

Q 13:- What is the significance of negative energy of the orbiting electron in an atom?
Ans:- The negative energy of an orbiting electron of an atom signifies that the electron is bound to the nucleus and the force between them is attractive in nature.

Q 14:- Why is the idea of stationary states very important in Bohr's atom model?
Ans:- Bohr atom is stable only on account of the concept of stationary state/orbit.

Q 15:- Bohr assumed that nucleus is fixed. Was this assumption correct? If not, what is the error in Bohr's predicted values for frequencies of the spectral lines?
Ans:- Truely speaking, nucleus of an atom is not fixed. When energy is emitted due to electronic transition from one energy state to another, there is recoil motion of nucleus too. However, as mass of nucleus is extremely large as compared to the mass of electron, the recoil speed of nucleus is very small and may be neglected. Therefore, practically, nucleus may be considered to be stationary.

Q 16:- How does atomic spectrum differ from a continuous spectrum?
Ans:- A continuous spectrum (e.g., spectrum of white sunlight) consists of lines of all possible wavelengths/frequencies, without any break, spread over a large region. However, an atomic gas or vapour, on proper excitation, emits radiation containing certain specific wavelengths only. Such a spectrum is termed as line atomic spectrum.

Read also: Class 12 Physics Chapter 12 MCQs with Answer Atoms with Answer

Q 17:- What is the difference between emission spectrum and absorption spectrum of an atomic gas?
Ans:- When an atomic gas is excited at low pressure, the emitted radiation is emission line spectrum which consists of bright lines on a dark background. Each bright line corresponds to a particular frequency/ wavelength.

When white light passes through an atomic gas and the transmitted light is analysed, some dark lines are observed in the spectrum. These dark lines correspond to those very wavelengths, which were found in the emission spectrum of the gas. This type of spectrum consisting of intermittant dark lines in white light spectrum is called the line absorption spectrum of the material of the gas.

Q 18:- What would happen if the electrons in an atom were stationary?
Ans:- If electrons in an atom were stationary, they would be pulled into the nucleus on account of coulombian attractive force. Such an atom would not be a stable atom.

Q 19:- How does an atom remain stable in which the electron is revolving around the nucleus in a definite orbit?
Ans:- If the electron is revolving around the nucleus in a definite orbit in the atom, coulombian force acting on the electron is utilised just to provide the requisite centripetal force and the electron continues to revolve in same orbit with a constant orbital speed. Thus, the atom will be stable provided that the electron is revolving in a nonradiating orbit (i.e., it does not emit radiation while revolving in its orbit.)

Q 20:- Is frequency of radiation emitted continuous or discrete as per (a) Rutherford atom model, and (b) Bohr atom model?
Ans:- (a) As per Rutherford atom model, the frequency of radiation emitted by an atom is continuous and exactly equal to the revolution frequency of the electron of the atom.

(b) As per Bohr atom model, the frequencies of radiation emitted are discrete. The radiation frequency is equal to the difference between the energies in two states (initial and final) of electron divided by Planck's constant h.

Q 21:- Although a hydrogen atom has only one electron yet its spectrum has many lines. Explain why.
Ans:- According to Bohr atom model, emission of a radiation photon of characteristic frequency takes place whenever the electron of hydrogen atom shifts from a higher energy excited state to another lower energy excited state or ground state. Depending upon the degree of excitation, so many transitions like (2-1), (3-2), (4-2), (4-1), etc., are possible. Each transition gives a spectral line of characteristic frequency. Thus, we obtain a large number of spectral lines in hydrogen spectrum irrespective of the fact that it has only one electron in an atom.

Q 22:- Why does Bohr atom model fail for multielectron atoms?
Ans:- As per Bohr atom model, we have considered electrostatic attractive force experienced by the electron due to positively charged nucleus only. For multielectron atoms, each electron will also experience repulsive forces due to other electrons present in the atom. As Bohr atom model does not take these repulsive forces into the account, it fails to explain atomic spectra of multielectron atoms.

Q 23:- 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?
Ans:- If one uses a thin sheet of solid hydrogen in place of the gold foil in an α-particle scattering experiment, no large angle scattering of α-particles will be possible. Mass of a hydrogen nucleus (proton) is only 1.67 x 10-27 kg which is even less than that of α-particle (ma = 6.7 x 10-27 kg). So, even in a head-on collision, the alpha particle will not bounce back. The situation is similar to a moving football colliding with a tennis ball at rest. In such a situation, one cannot expect football to bounce back without disturbing the tennis ball.

Q 24:- Answer the following questions, which help you understand the difference between Thomson's model and Rutherford's model better.

(a) Is the average angle of deflection of α-particles by a thin gold foil predicted by Thomson's model much less, about the same, or much greater than that predicted by Rutherford's model?

(b) Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson's model much less, about the same, or much greater than that predicted by Rutherford's model?

(c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?

(d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of α-particles by a thin foil?
Ans:- (a) The average angle of deflection (θ) of a-particles by a thin gold foil as predicted by Thomson's model is about the same that predicted by as Rutherford's atom model.

(b) The probability of backward scattering of a-particles as predicted by Thomson's model is much less than that predicted by Rutherford's model.

(c) Linear dependence of the number of a-particles scattered at moderate angles on thickness t of gold foil suggests that scattering is predominantly due to a single collision. Chances of a single collision increase linearly with the number of target atoms and hence, linearly with the thickness of gold foil.

(d) In Thomson model, positive charge is uniformly distributed in entire atom. So, a single collision causes very little deflection and hence, average scattering can be explained only on the basis of multiple scattering. Therefore, it is completely wrong to ignore multiple scattering in Thomson model. In Rutherford model, whole positive charge is concentrated in nucleus and most of the deflection comes from a single scattering. So, multiple scattering may be ignored.

Q 25:- If Bohr's quantisation postulate (angular momentum `=\frac{nh}{2\pi}` is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?
Ans:- If one applies Bohr's quantisation principle for planetary motion (say, Earth), then value of principal quantum number n = 1070. For such large values of n, the difference in angular momenta and energy between successive levels is meaningless and we may consider angular momenta as well as energy levels as continuous ones.

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