Atoms Class 12 Notes: CBSE Physics Chapter 12

Rutherford’s Alpha Scattering Experiment:

Following are the observations made through this experiment:

1. Most of  $\alpha $ -particles were seen to pass through the fold foil without any appreciable deflection. 

2. The various $\alpha $ particles, on passing through the gold foil, undergo different amounts of deflections. A large number of  $\alpha $ particles caused fairly large deflections. 

3. A very small number of  $\alpha $ -particles (about \[1\text{ }in\text{ }8000\] ) practically retracted their paths or suffered deflection of nearly\[180{}^\text{o}\] . 

4. The plot between the total number of  $\alpha $- particles \[N\left( \theta  \right)\] scattered through angle $\theta $ and the scattering angle $\theta $  was found to be as shown in graph:

These experimental observations led Rutherford to the following inferences:

1. As most of the $\alpha $ particles passed without any deviation, the atom had a lot of empty space in it. 

2. As fast and heavy $\alpha $ particles could be deflected even through\[180{}^\text{o}\], the whole of the positive charge and practically the entire mass of the atom were confirmed to extremely small central cores. These were called nuclei. Because $1$ in about $8000$ $\alpha $ particles get deflected through ${{180}^{{}^\circ }}$, the size of the nucleus was assumed to be about $1/10000th$ of the size of the atom.

8.1 Rutherford’s Atom Model 

On the basis of the results of $\alpha $ scattering experiment, Rutherford suggested the following picture of an atom: 

1. Atoms can be regarded as spheres of diameters \[{{10}^{10}}m\] but whole of the positive charge and almost the entire mass of these atoms are concentrated in small central cores called nuclei having diameters of about \[{{10}^{14}}m.~\]

2. The nucleus is neighbored by electrons. In other words, the electrons are distributed over the remaining part of the atom leaving plenty of empty space in the atom.

Drawbacks of Rutherford’s Atom Model:

1. When the electrons revolve around the nucleus, they get continuously accelerated towards the centre of the nucleus. According to Lorentz, an accelerated charged particle must radiate energy continuously. Thus, in the atom, a revolving electron must continuously emit energy and hence the radius of its path must go on decreasing and finally, it must fall into the nucleus. However, electrons revolve around the nucleus without falling into it. Clearly, Rutherford’s atom model couldn’t explain the stability of the atom. 

2. Suppose if Rutherford's atom model is true, the electron could revolve in orbits of all possible radii and thus it should emit a continuous energy spectrum. But, atoms like hydrogen possess a line spectrum. 

Alpha-particle Trajectory:

• The path of an α-particle depends on the impact parameter, which is the distance from the particle’s starting point to the centre of the nucleus.

• In a beam of α-particles, each particle has a different impact parameter, causing them to scatter in various directions. They all have nearly the same energy.

• An α-particle that comes very close to the nucleus (small impact parameter) scatters a lot. If it hits the nucleus head-on (very small impact parameter), it bounces back (θ ≈ 180°).

• If the impact parameter is large, the α-particle barely changes direction (θ ≈ 0°).

• The fact that few α-particles bounce back means that head-on collisions are rare, showing that most of the atom's mass and positive charge is concentrated in a small area. This helps determine the size of the nucleus.

Electron Orbits (Rutherford Model):

1. Rutherford’s Nuclear Model:

The atom is modelled as a small, dense, positively charged nucleus with electrons revolving around it in fixed orbits.

The force of attraction between the nucleus and the electrons provides the necessary centripetal force to keep the electrons in stable orbits.

2. Centripetal Force:

The electrostatic force ( $F_e$ ) between the electron and the nucleus equals the centripetal force ($ F_c $):

$   F_e = F_c = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{r^2} = \frac{m v^2}{r}$

where:

$\epsilon_0 $ = permittivity of free space

e = electron charge

r = orbit radius

m = electron mass

 v = velocity of the electron

3. Relation between Orbit Radius and Electron Velocity:

The radius r of the electron's orbit is related to its velocity v by the formula:

 $   r = \frac{e^2}{4 \pi \epsilon_0 m v^2}$

4. Kinetic Energy (K):

The kinetic energy K of the electron is given by:

$   K = \frac{1}{2} m v^2 = \frac{e^2}{8 \pi \epsilon_0 r}$

5. Electrostatic Potential Energy (U):

The potential energy U of the electron due to the nucleus is:

 $U = -\frac{e^2}{4 \pi \epsilon_0 r}$

The negative sign indicates that the force is attractive.

6. Total Energy (E):

The total energy E of the electron in the hydrogen atom is the sum of its kinetic and potential energy:

$E = K + U = \frac{e^2}{8 \pi \epsilon_0 r} - \frac{e^2}{4 \pi \epsilon_0 r}$

$   E = -\frac{e^2}{8 \pi \epsilon_0 r}$

The total energy is negative, meaning the electron is bound to the nucleus.

Atomic Spectra:

1. Characteristic Spectrum of Elements:

Each element has its own unique spectrum of radiation, which it emits.

2. Emission Line Spectrum:

• When a gas or vapour is excited at low pressure (often by an electric current), it emits radiation with specific wavelengths.

• This is called the emission line spectrum, where bright lines appear on a dark background.

• The spectrum of atomic hydrogen is an example.

3. Identification through Emission Line Spectra:

The study of an element's emission line spectrum can be used like a "fingerprint" to identify the gas.

4. Absorption Spectrum:

• When white light passes through the gas, some wavelengths are absorbed, creating dark lines in the spectrum.

• These dark lines match the wavelengths in the emission line spectrum of the gas.

• This is called the absorption spectrum.

Bohr Atomic Model: 

Bohr considered the Rutherford model of the atom and added a few arbitrary conditions in it. These conditions are referred to as his postulates:

1. The electron in a stable orbit does not radiate energy. i.e., \[\frac{m{{v}^{2}}}{r}=\frac{kZ{{e}^{2}}}{{{r}^{2}}}\].

2. A stable orbit is that in which the angular momentum of the electron about nucleus is an integral \[\left( n \right)~\] multiple of\[\frac{h}{2\pi }~i.e.,\text{ }mvr~=n\frac{h}{2\pi };~n=1,2,~3,.......\left( n\ne 0 \right)\]. 

3. The electron can absorb or radiate energy only when the electron jumps from a lower to a higher orbit or falls from a higher to a lower orbit.

4. The energy emitted or absorbed is a light photon of frequency $\nu $ and of energy, \[E~=~h\nu \].

10.1 For Hydrogen Atom: \[\left( Z=atomic\text{ }number=1 \right)\]:

1. \[{{L}_{n}}=\] angular momentum in the ${{n}^{th}}$ orbit $=n\frac{h}{2\pi }$ 

2. \[{{r}_{n}}=\] radius of \[{{n}^{th}}\] circular orbit \[=\left( 0.529\text{ }A{}^\text{o} \right){{n}^{2}}\] \[\left( 1A{}^\text{o}={{10}^{-10}}m \right)\]

\[\Rightarrow {{r}_{n}}\propto {{n}^{2}}\]

3. \[{{E}_{n}}=\] Energy of the electron in the\[~{{n}^{th}}~\]orbit $=\frac{-13.6eV}{{{n}^{2}}}$

$\Rightarrow {{E}_{n}}\propto \frac{1}{{{n}^{2}}}$ 

Note: 

Total energy of the electron in an atom is negative, mentioning that it is bound.

Binding energy \[{{\left( B.E \right)}_{n}}=-{{E}_{n}}=\frac{13.6V}{{{n}^{2}}}\]

4. ${{E}_{{{n}_{2}}}}-{{E}_{{{n}_{1}}}}=$ Energy emitted when an electron jumps from ${{n}_{2}}^{th}$ orbit to ${{n}^{th}}_{1}$ orbit $\left( {{n}_{2}}>{{n}_{1}} \right)$ 

$\Rightarrow \Delta E=\left( 13.6ev \right)\left[ \frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}^{2}}_{2}} \right]$ 

$\Rightarrow \Delta E=h\nu $ 

where, 

$\nu =$frequency of spectral line emitted.

$\frac{1}{\lambda }=\nu =$ wave number [number of waves in unit length]

$\Rightarrow \nu =R\left[ \frac{1}{{{n}^{2}}_{1}}-\frac{1}{{{n}^{2}}_{2}} \right]$ 

where, $R=$ Rydberg’s constant for hydrogen$=1.097\times {{10}^{7}}{{m}^{-1}}$ 

5. For hydrogen like atom/species of atomic number $Z$:

${{r}_{nz}}=\frac{Bohr\text{ }radius}{Z}{{n}^{2}}=\left( 0.529{{A}^{o}} \right)\frac{{{n}^{2}}}{Z}$ 

${{E}_{nz}}=\left( -13.6 \right)\frac{{{Z}^{2}}}{{{n}^{2}}}ev$ 

${{R}_{z}}=R{{Z}^{2}}-$ Rydberg’s constant for the element of atomic number Z.

Note: When motion of the nucleus is also considered, then $m$ is replaced by $\mu $, where

$\mu =$ reduced mass of electron-nucleus system $=\frac{mM}{\left( m+M \right)}$

In this case, ${{E}_{n}}=\left( -13.6ev \right)\frac{{{Z}^{2}}}{{{n}^{2}}}\frac{\mu }{{{m}_{e}}}$

Energy levels

1. Energy in Atom:

The energy of an atom is lowest (most negative) when the electron is closest to the nucleus. This happens when n = 1.

For higher values of n (like 2, 3, ...), the energy is larger, meaning the electron has more energy in outer orbits.

2. Ground State:

The lowest energy state, called the ground state, occurs when the electron is in the smallest orbit (Bohr radius, $a_0$ with energy $E_1 = -13.6 $eV).

To remove the electron from this state (free it), the energy required is 13.6 eV, called ionisation energy of hydrogen.

3. Excited States:

• At room temperature, most hydrogen atoms are in their ground state.

• When the atom gains energy (e.g., through collisions), the electron can jump to a higher energy level, entering an excited state.

• For n = 2, the energy is $E_2 = -3.40 \text{ eV}$. To move from n = 1 to n = 2, the energy required is $10.2 \text{ eV}$.

4. Higher Excited States:

• For n = 3, the energy is $E_3 = -1.51 \text{ eV}$, and moving from n = 1 to n = 3 requires $12.09 \text{ eV}$.

•  As n increases, the energy levels become closer together, requiring less energy to move between them.

5. Energy Level Diagram:

• The highest energy state corresponds to n =$ \infty$, where the energy is 0 eV, meaning the electron is completely free from the nucleus.

• As n increases, the excited state energies come closer together.

The Line Spectra of the Hydrogen Atom

1. Bohr's Third Postulate:

When an atom transitions from a higher energy state ($ n_i$)) to a lower energy state ($ n_f $), the energy difference is released as a photon.

The frequency ($ \nu_{if} $) of the emitted photon is related to the energy difference:

 $   h \nu_{if} = E_{ni} - E_{nf}$

where h is Planck's constant.

2. Energy Levels of Hydrogen Atom:

The electron in a hydrogen atom usually stays in the ground state at room temperature.

To remove an electron from the ground state (ionisation), 13.6 eV of energy is needed.

Above E = 0, the electron is free, meaning there are infinite possible energy states.

3. Emission Lines:

When an electron moves from a higher to a lower energy level, it emits a photon. This creates spectral lines called emission lines.

4. Absorption Lines:

If an atom absorbs a photon with exactly the right energy, the electron moves to a higher energy state. This creates absorption lines in the spectrum.

5. Spectral Analysis:

When light passes through a gas, dark absorption lines appear in the spectrum at the frequencies absorbed by the atoms.

6. Bohr's Contribution:

Bohr's model of the hydrogen atom helped explain atomic spectra and was a major step toward quantum theory. He won the Nobel Prize in Physics in 1922 for this work.

Bohr’s Second Postulate and de Broglie’s Explanation:

1. Bohr's Second Postulate:

Bohr proposed that the angular momentum of an electron in orbit is quantized. The angular momentum can only take specific values:  

$   L_n = \frac{nh}{2\pi}, \quad n = 1, 2, 3, \ldots$

2. Why is Angular Momentum Quantized?:

In 1923, Louis de Broglie explained this puzzle by proposing that electrons, like waves, have wave-like properties. This was 10 years after Bohr’s model.

In Chapter 11, we learned that material particles (like electrons) have a wave nature, verified experimentally in 1927 by C.J. Davisson and L.H. Germer.

3. Wave Nature of Electrons:

de Broglie suggested that electrons moving in circular orbits should be viewed as particle waves, similar to standing waves on a string.

4. Standing Waves on a String:

Only specific wavelengths form standing waves when a string is plucked. Waves with different wavelengths interfere with themselves and disappear.

For an electron in the nth orbit of radius \(r_n\), the total distance the wave travels is the orbit's circumference:

$   2\pi r_n = n\lambda, \quad n = 1, 2, 3, \ldots$

5. Electron Momentum and de Broglie Wavelength:

The de Broglie wavelength of an electron is given by:

$   \lambda = \frac{h}{p}$

For an electron with speed $v_n$, the momentum is $mv_n$, so:

 $   \lambda = \frac{h}{mv_n}$

6. Bohr's Quantum Condition:

From the above equation, the quantum condition for angular momentum is derived as:

$   m v_n r_n = \frac{nh}{2\pi}$

Limitations of Bohr's Model:

1. Applicable to Hydrogenic Atoms Only:

The Bohr model works for atoms with only one electron, like hydrogen, but not for multi-electron atoms like helium. This is because electrons also interact with each other, which Bohr's model doesn't account for.

2. Unable to Explain Relative Intensities:

The model correctly predicts the frequencies of light emitted by hydrogen atoms but cannot explain why some frequencies are stronger or weaker in intensity.

3. Quantum Mechanics Required:

For more complex atoms, a new theory called Quantum Mechanics is needed, which provides a more complete understanding of atomic structure.

Download Chapter 12 Atoms Revision Notes

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Q1: Determine the shortest wavelength existing in the Paschen series of spectral lines.

Ans: 

Using Rydberg’s formula here:

hc/λ= 2.18 x 10-19 \[\left \lceil \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right \rceil\] ---(1) 

Where h = 6.626 x 10-34Js

            c = 3 x 108m/s

For the Paschen series, n1 = 3,  n2 = ∞

Putting these values in eq (1):

(6.626 x 10-34 x 3 x 108/λ) =21.8 x 10-19 \[\left \lceil \frac{1}{3^{2}} - \frac{1}{\infty^{2}} \right \rceil\]

On solving, we get λ as;

              λ = 820 nm

Q2: Total energy of an electron in the first excited state of the H-atom is 3.4 eV. Find the following:

1. Kinetic energy (KE) of e- in this state

2. Potential energy (PE) of e- this state

3. Which of the above answers would vary if the alternative of zero potential energy is altered?

Ans: 

1. KE = - (Total energy in the first state) = - (- 3.4 eV) = + 3.4 eV

2. PE = Negative of twice of the KE

                 = - (2 KE) = - (2 x 3.4) = - 6.8 eV

3. PE of a system depends on the reference point. Here, we have taken the reference point as zero. If this point is changed, then PE also changes, which means the total energy (KE + PE) of a system will also change.

Atoms Class 12 Notes Physics - Basic Subjective Questions

Section-A (1 Mark Questions)

1. Why Thomson's model of the atom is known as a plum pudding model? 

Ans. This is because in Thomson's model the electrons are assumed to be uniformly embedded in a sphere of positively charged matter like the plums are arranged in a pudding. 

2. Why did Thomson's atomic model fail? 

Ans. Thomson model failed to explain the scattering of alpha-Particles through large angles in Rutherford’s experiment. 

3. Write two important inferences drawn from Rutherford's alpha particle scattering experiment.

Ans.

(i) Mass is concentrated in a very small volume of the atom, called nucleus. 

(ii) The nuclear radius is about 1/10,000 of the atomic radius. 

4. Why do we use a very thin gold foil in Rutherford's α-particle scattering experiment? 

Ans. In thick foil, the entire kinetic energy of the α-particles will be absorbed and so particles will not be able to penetrate through the foil. 

5. What is the significance of the negative energy of electron in the orbit? 

Ans. This signifies that the electron is bound to the nucleus. Due to electrostatic attraction between electron and nucleus, the P.E. is negative and is greater than K.E. of electron. Total energy of electron is negative so it cannot escape from the atom. 

6. What are stationary orbits? 

Ans. Bohr postulated that electrons can revolve around the nucleus in certain discrete, non-radiating orbits in which the angular momentum of an electron is an integral multiple of h/2π. Such orbits are called stationary orbits. 

7. State Bohr's quantization condition in terms of de-Broglie wavelength. 

Ans. According to Bohr's quantization condition, only such circular orbits are allowed for stationary states of an electron which contain an integral multiple of de-Broglie wavelengths. 

8. How much energy is possessed by an electron for n=∞ (hydrogen atom)? 

Ans. Zero, For $n=\infty ,E_{n}=-\dfrac{13\cdot 6}{n^{2}}eV=0$

9. In a hydrogen atom, if the electron is replaced by a particle which is 200 times heavier but has the same charge, how would its radius change? 

Ans. Radius, $r\varpropto\dfrac{1}{m}$

When the electron is replaced by a 200 times heavier particle, the radius decreases to $\dfrac{1}{200}$of the original radius.

10. The short wavelength limits of the Lyman, Paschen and Balmer series, in the hydrogen spectrum, are denoted by $\lambda _{L},\lambda _{P}$ and $\lambda _{B}$ respectively. Arrange these wavelengths in increasing order. 

Ans. $\lambda _{L}< \lambda _{B}< \lambda _{P}$

Section-B (2 Marks Questions)

11. Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level. 

Ans. Energy of photon,  $\therefore \dfrac{(hv)_{2\rightarrow 1}}{(hv)_{\infty \rightarrow 1}}=\dfrac{\left ( \dfrac{1}{1^{2}}-\dfrac{1}{2^{2}} \right )}{\left ( \dfrac{1}{1^{2}}-\dfrac{1}{\infty ^{2}} \right )}=\dfrac{3}{4}=3:4$

12. In the Rutherford scattering experiment, the distance of closest approach for an a-particle is do. If α-particle is replaced by a proton, then how much kinetic energy in comparison to a particle will be required to have the same distance of closest approach ? 

Ans. When $\alpha -$particle is replaced by proton, then there will be change in atomic number and mass of the particle.

Distance of closest approach $d_{0}=\dfrac{2kZe^{2}}{k_{x}}$ (where, $k=\dfrac{1}{4\pi \varepsilon _{0}}$ )

$\vec{}$ for given distance of closest approach, kinetic energy $\varpropto\;Z$ (atomic number)

$\Rightarrow \dfrac{K_{proton}}{K_{\alpha }}=\dfrac{Z_{proton}}{Z_{\alpha }}=\dfrac{1}{2}$

$\Rightarrow K_{proton}:K_{\alpha }=1:2$

13. The radius of innermost electron orbit of a hydrogen atom is $5\cdot 3\times 10^{-11}$ m. What is the radius of orbit in the second excited state? 

Ans. The radius of whose principal quantum number is n is given by $r=n^{2}r_{0}$

Where, r0 = radius of innermost electron orbit for hydrogen atom, $r_{0}=5\cdot 3\times 10^{-11}$ m

For second excited state, n = 3

$\therefore r=3^{2}\times r_{0}=9\times 5\cdot 3\times 10^{-11}$

$r=4\cdot 77\times 10^{-10}m$

14. Using Bohr’s postulates of the atomic model, derive the expression for radius of nth electron orbit, thus obtaining the expression for Bohr’s radius. 

Ans. Form de-Broglie hypothesis, wavelength associated with electron

$\lambda =\dfrac{h}{mv}$

$mv=\dfrac{h}{\lambda }$

From Bohr’s postulate, $mvr=n\dfrac{h}{2\pi }$

We get, $\dfrac{h}{\lambda }r=n\dfrac{h}{2\pi }$

$2\pi r=n\lambda$

i.e., circumference $(S=2\pi r)$ of nth permitted orbit for the electron can contain exactly n wavelength of de-Broglie wavelength associated with electron in that orbit.

15. Total energy of an electron in an atom is negative. What does it signify ? Calculate the energy required to move a electron from ground state to first excited state. 

Ans. The negative sign indicates that the electron is bound to the nucleus by electrostatics forces of attraction. Energy in ground state, E1 =−13.6eV

Energy in first excited state, $E^{2}=\dfrac{13\cdot 6}{2^{2}}=-3\cdot 4eV$ required energy, $\Delta E=\Delta E_{2}-E_{1}=-3\cdot 4+13\cdot 6=10\cdot 2eV$

5 Important Formulas of Physics Class 12 Chapter 12 Atoms

Importance of Physics Chapter 12 Atoms Physics Class 12 Notes PDF

• Revision notes help us quickly understand and remember key concepts before exams.

• They save time by focusing on essential information and skipping unnecessary details.

• They provide practical examples that show how theoretical knowledge is used in real-life situations.

• Revision notes ensure thorough preparation by covering all important topics in a structured manner.

• They increase confidence by clearly understanding what to expect in exams.

• Accessible formats like PDFs allow for easy studying anytime and anywhere.

Tips for Learning the Class 12 Physics Atoms Notes

• Start by understanding the basics of atomic models, such as Bohr’s model. Focus on how it explains the structure of atoms and the quantization of energy levels.

• Familiarise yourself with important formulas like Bohr’s radius, energy of orbits, and the Rydberg formula. Practice using these formulas in different problems to strengthen your understanding.

• Understand the significance of quantum numbers (principal, azimuthal, magnetic, and spin) in defining electron positions and energies. Relate these to atomic models discussed in the chapter.

• Study the production of atomic spectra, including emission and absorption spectra. Learn how these spectra are related to electron transitions between energy levels.

• Familiarise yourself with de Broglie's concept of wave-particle duality. Practice calculating de Broglie wavelengths for particles like electrons.

• Relate theoretical concepts to real-life experiments and observations, such as the Balmer series in the hydrogen spectrum. This connection reinforces your understanding.

Conclusion

Vedantu's revision notes for Class 12 Physics Chapter 12, "Atoms," provide a clear and concise review of atomic theory. You’ve explored essential topics like atomic models, energy levels, and spectral lines. These notes simplify complex concepts and formulas, helping you understand and apply them effectively. By reviewing the key points and practising problems, you’ll be well-prepared for exams. Use these notes to improve your knowledge and clarify any doubts. With a solid grasp of the chapter, you’ll be able to tackle questions confidently and achieve better results. Keep revising and practising to ensure a strong grasp of atomic physics.

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