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NCERT Class 12 Physics Textbook Chapter 12 With Answer PDF Free Download
Chapter 12: Atoms
12.1 INTRODUCTION
By the nineteenth century, enough evidence had accumulated in favor of the atomic hypothesis of matter.
In 1897, the experiments on electric discharge through gases were carried out by the English physicist J. J. Thomson (1856 – 1940) revealed that atoms of different elements contain negatively charged constituents (electrons) that are identical for all atoms.
However, atoms on a whole are electrically neutral. Therefore, an atom must also contain some
positive charge to neutralize the negative charge of the electrons.
But what are the arrangement of the positive charge and the electrons inside the atom? In other words, what is the structure of an atom?
The first model of the atom was proposed by J. J. Thomson in 1898.
According to this model, the positive charge of the atom is uniformly distributed throughout the volume of the atom and the negatively charged electrons are embedded in it like seeds in a watermelon.
This model was picturesquely called the plum pudding model of the atom.
This radiation is considered to be due to oscillations of atoms and molecules, governed by the interaction of each atom or molecule with its neighbors.
In contrast, light emitted from rarefied gases heated in a flame, or excited electrically in a glow tube such as the familiar neon sign or mercury vapor light has only certain discrete wavelengths.
The spectrum appears as a series of bright lines. In such cases, the average spacing between atoms is large.
Hence, the radiation emitted can be considered due to individual atoms rather than because of interactions between atoms or molecules.
In the early nineteenth century, it was also established that each element is associated with a characteristic spectrum of radiation, for example, hydrogen always gives a set of lines with a fixed relative position between the lines.
This fact suggested an intimate relationship between the internal structure of an atom and the spectrum of radiation emitted by it.
In 1885, Johann Jakob Balmer (1825 – 1898) obtained a simple empirical formula that gave the wavelengths of a group of lines emitted by atomic hydrogen.
Since hydrogen is the simplest of the elements known, we shall consider its spectrum in detail in this chapter.
Alpha-particle trajectory
The trajectory traced by an α-particle depends on the impact parameter, b of collision. The impact parameter is the perpendicular distance of the initial velocity vector of the α-particle from the center of the nucleus (Fig. 12.4).
A given beam of α-particles has a distribution of impact parameters b so that the beam is scattered in various directions with different probabilities (Fig. 12.4).
(In a beam, all particles have nearly the same kinetic energy.) It is seen that an α-particle close to the nucleus (small impact parameter) suffers large scattering.
In case of a head-on collision, the impact parameter is minimum and the α-particle rebounds back (θ ≅ π). For a large impact parameter, the α-particle goes nearly undefeated and has a small deflection (θ ≅ 0).
| Author | NCERT |
| Language | English |
| No. of Pages | 24 |
| PDF Size | 412 KB |
| Category | Physics |
| Source/Credits | ncert.nic.in |
NCERT Solutions Class 12 Physics Chapter 12 Atoms
Q1: Choose the correct alternative from the clues given at the end of 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.)
Ans:
(a) The size of the atom in Thomson’s model is no different from the atomic size in Rutherford’s model.
(b) In the ground state of Thomson’s model electrons are in stable equilibrium, while in Rutherford’s model 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 a 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 models.
Q2: 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:
We know that the mass of the incident alpha particle (6.64 × 10-27kg) is more than the mass of hydrogen (1.67 × 10-27Kg). Hence, the target nucleus is lighter, from which we can conclude that the alpha particle would not rebound. Implying the fact that solid hydrogen isn’t a suitable replacement for gold foil for the alpha particle scattering experiment.
Q3: What is the shortest wavelength present in the Paschen series of spectral lines?
Ans:
We know the Rydberg’s formula is written as:\frac{hc}{\lambda }=21.76\times 10^{-19}[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}]λhc=21.76×10−19[n121−n221]
Where,
h = Planck’s constant = 6.6 × 10-34
c = Speed of light = 3 × 108 m/s
n1 and n2 are integers
So the shortest wavelength present in the Paschen series of the spectral lines is given for n1 = 3 and n2 = ∞\\\frac{hc}{\lambda }=21.76\times 10^{-19}[\frac{1}{3^{2}}-\frac{1}{\infty^{2}}] \\ \\\lambda =\frac{6.6\times 10^{-34}\times 3\times 10^{8}\times 9}{21.76\times 10^{-19}}λhc=21.76×10−19[321−∞21]λ=21.76×10−196.6×10−34×3×108×9
= 8.189 × 107 m
= 818.9 nm
Q4: 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?
Ans:
It is given that
Separation of two energy levels in an atom,
E = 2.3 eV = 2.3 × 1.6 × 10-19
= 3.68 × 10-19 J
Consider v as the frequency of radiation emitted when the atom transits from the upper level to the lower level.
So the relation for energy can be written as:
E = hv
Where,
h = Planck’s constant = 6.62 × 10-34 Js
v = E/h
= \frac{3.68 \times 10^{-19}}{6.62\times 10^{-32}}6.62×10−323.68×10−19
= 5.55 × 1014 Hz
= 5.6 × 1014 Hz
Therefore, the frequency of radiation is 5.6 × 1014 Hz.
NCERT Class 12 Physics Textbook Chapter 12 With Answer PDF Free Download