Atoms: Thomson's Atomic Model, Rutherford's Model and Atomic Spectra

Collegedunia Team logo

Collegedunia Team Content Curator

Content Curator

Everything we see around us, from a tiny grain of sand to a huge star in the sky, is made up of tiny particles called atoms. It is the basic building block of all matters.

  • An atom is a particle composed of a nucleus of protons and neutrons surrounded by electrons.
  • They can combine with other atoms to form molecules.
  • An atom consists of subatomic particles: electrons, protons, and neutrons.
  • The protons and neutrons form the nucleus of an atom.
  • A neutral atom contains an equal number of protons and electrons.
  • So many physicists tried to explain the structure of atoms.
  • But after that, it was Neils Bohr whose explanation about the structure was well accepted.

Key Terms: Atoms, Thomson's Atomic Model, Rutherford's Model, Atomic Spectra, Alpha-Particle trajectory, Electrons, Protons, Neutrons, Atom, Impact parameter


Thomson’s Atomic Model

[Click Here for Sample Questions]

There are many different models of an atom we will learn in this article, but among them, the first model of the atom was proposed by J. J. Thomson. He proposed that

  • Each atom consists of negatively charged particles called electrons.
  • Since the atom as a whole is neutral, there must be positively charged particles in the atom.
  • An atom is a sphere of positive charges of uniform density in which negative charges are embedded like plums in the pudding.
  • Thomson's model of the atom is also called the “Plum-pudding model”.
  • Thomson assumed that the harmonic oscillations of the electrons about their mean position cause the emission of light.
  • He predicted if a beam of alpha particles passed through an atom, it would suffer only small deflection due to weak electric forces exerted on it by positive charges.

Also Read:


Alpha Particle Scattering Experiment

[Click Here for Previous Year Questions]

As we know alpha particles are doubly positively charged helium nuclei, an experiment was carried out by Ernst Rutherford along with his associates Hans Geiger and Ernst Marsden on the scattering of α-particle due to atoms of thin gold foils.

The conclusion from this experiment was obtained as

  • As most of the α-particle passes through gold foil undeflected, it indicates that most of the space in an atom is empty.
  • Since α-particles are positively charged large mass could only be deflected backward by a large repulsive force by a heavily positively charged region.
  • He concluded that positive charges in an atom were concentrated in a very small region at the center of an atom called a nucleus.
  • The α-particle that travels toward the nucleus directly moves in the opposite direction due to the large repulsive force between the α-particle and the nucleus.
  • Electrons being very light do not affect the motion of the α-particles.
Alpha particle scattering experiment
Alpha particle scattering experiment

Rutherford's Alpha Ray Scattering Experiment Detailed Video Explanation:


Rutherford’s Model

[Click Here for Sample Questions]

Based on the results of the alpha particle scattering experiment, Rutherford proposed a new model of the atom. According to this model

  • Almost all of the mass of the atom and all the positive charges of an atom are concentrated in a very small region called the nucleus of the atom.
  • The size of the nucleus is extremely small (radius 10-14 to 10-15 m) as compared to the size of the atom.
  • The negatively charged particles known as electrons revolve around the nucleus.
  • So most of the space in an atom is empty.
  • The centripetal force required for revolving the electrons around the nucleus is provided by Coulomb’s force of attraction between the nucleus and the electrons.
  • The number of revolving electrons is equal to the number of positive charges in the nucleus.
  • Hence atom is electrically neutral.

Atoms Class 12 Notes PDF


Alpha-Particle Trajectory and Impact Parameter

[Click Here for Previous Year Questions]

The perpendicular distance of the velocity vector from the center of the nucleus is known as the Impact parameter (b).

The angle between the direction of approach of the alpha-particle and the direction of the scattered alpha-particle is called the Scattering angle (θ).

The expression of impact parameter derived by Rutherford in terms of scattering angle θ is given by

\(b=\frac {Ze^2cot(\frac {\theta}{2})}{4 \pi \epsilon_o(\frac {1}{2}mv^2)}\)

Alpha particle trajectory
Alpha particle trajectory

Conclusions:

  • For a large value of b, cot θ/2 is large, and hence scattering angle θ will be small i.e. alpha particles traveling far away from the nucleus suffer small deflection.
  • If the value of b is small, cot θ/2 is also small, and hence scattering angle θ will be large i.e. alpha particles traveling close to the nucleus suffer large deflection.
  • When b = 0 i.e. the alpha particle is directed toward the center then cot θ/2 = 0 or θ = 180. Thus the alpha-particle traveling towards the center of a nucleus retraces its path.

Atomic Spectra

[Click Here for Sample Questions]

When an atomic gas or vapor is excited at lower pressure, the emitted radiation has a spectrum that contains certain specific wavelengths only.

  • A spectrum of this kind is termed an emission line spectrum and it consists of bright lines on dark backgrounds.
  • The study of emission line spectra of material can serve as a type of fingerprint for the identification of the gas.
  • When white light passes through gas and we analyze the transmitted light using a spectrometer we find some dark lines in the spectrum.
  • The wavelengths that were found in the emission line spectrum of the gas and the dark lines correspond precisely to those wavelengths.
  • This is called the absorption spectrum of the gas.

Spectral Series

Hydrogen has the simplest spectrum. In the observed spectrum the first sight there does not seem to be any regularity in spectral lines.

  • However, the spacing between lines of the hydrogen spectrum decreases regularly.
  • Each of these sets is called a spectral series.
  • Balmer observed such a series in the visible region of the hydrogen spectrum.
  • This series is called the Balmer series.

In the spectral series,

  • The line with the longest wavelength, 656.3 nm in the red is called Hα
  • The next line with a wavelength of 486.1 nm in the blue-green is called Hβ
  • In the spectral series the third line with a wavelength of 434.1 nm in violet and is called Hγ; and so on.
Spectral lines of hydrogen atom
Spectral lines of hydrogen atom

Balmer gave an empirical formula for the wavelength of the different lines in the spectral series of the hydrogen atom. It is given by

\(\frac {1}{\lambda}=R(\frac {1}{2^2}-\frac {1}{n^2})\)

Where

  • λ is the emitted wavelength
  • n = 3, 4, 5, …..
  • R is Rydberg constant = 1.097 x 107 m-1

This equation is called the Balmer formula.

Other series of spectra for hydrogen were known, like Lyman, Paschen, Brackett, and Pfund series. These are

  • Lyman series: The spectral lines emitted due to the transition of an electron from any higher orbit (ni = 2, 3, 4, 5, …..) to the first orbit (nf = 1) form a spectral series known as the Lyman series.
  • Paschen series: The spectral lines emitted due to the transition of an electron from any higher orbit (ni = 4, 5, 6,…..) to the first orbit (nf = 3) form a spectral series known as the Paschen series.
  • Brackett series: The spectral lines emitted due to the transition of an electron from any higher orbit (ni = 5, 6, 7 …..) to the first orbit (nf = 4) form a spectral series known as the Brackett series.
  • Pfund series: The spectral lines emitted due to the transition of an electron from any higher orbit (ni = 6, 7, 8, …..) to the first orbit (nf = 5) form a spectral series known as the Pfund series.

Spectral series

Spectral series

Bohr’s Model of Hydrogen Atom

[Click Here for Previous Year Questions]

The postulates of Bohr’s Model of Hydrogen Atoms are given by

  • The electrons of the hydrogen atom revolve around in stable orbits without emitting any radiant energy.
  • In the stationary orbit of the atom, the angular momentum is the multiple of the equation h/2π and L = nh/2π, where n is called the quantum number.
  • The electron of the hydrogen makes the transition from a non-radiating orbit to another lower energy orbit, when the transition occurs a photon gets emitted.
  • The energy of a photon is the same as the energy difference between the final and the initial states. The frequency is h= Ei - Ef

Energy Level of a Hydrogen Atom

The energy of the hydrogen atom in the nth energy state is given by

\(E_n=-\frac{13.6}{n^2}eV\)

  • When n = 1, then E1 = -13.6 eV. This is the energy of the ground state of a hydrogen atom.
  • When n = 2, then E2 = -3.4 eV. This is the energy of the first excited state of a hydrogen atom.
  • When n = 3, then E3 = -1.51 eV. This is the energy of the second excited state of a hydrogen atom.
  • When n = ∞, then E∞​ = 0 eV. This is the energy of the highest energy state of a hydrogen atom.
Energy level of Hydrogen atom
Energy state of a hydrogen atom

Explanation of De Broglie’sTheory

[Click Here for Sample Questions]

Einstein suggested that light behaves both as a material particle as well as a wave. De-Broglie extended Einstein’s view and he said that all forms of matter like electrons, protons, neutrons, etc also have a dual character.

He further said that wavelength associated with a particle of mass “m” moving with velocity “v” is given by –

λ = h/mv

where λ is called de- Broglie wavelength

An electron behaves as a standing wave or stationary wave.

  • This wave extends around the nuclei in a circular orbit.
  • The electron wave is said to be in phase if the two ends of the electron wave meet to give a regular series i.e. there is constructive interference of electron waves and the electron motion has a character of a standing wave or non-energy radiation motion.

Therefore,

λ=2πr/n

Where n denotes the number of wavelengths associated with an electron wave.

From these equations, we can write-

mvr=nh/2π

The total energy of the electron remains constant because an electron revolving in a permitted orbit does not radiate energy.

Also Read:

Important Topics for JEE Main

As per JEE Main;2024 Session 1, important topics included in the chapter Atoms are as follows:

Topics Number of Questions Asked
Rutherford’S Nuclear Model Of Atom 1
Bohr Model Of The Hydrogen Atom 4

Some important memory based questions from JEE Main 2024 Session 1 are given below:

1. Given below are two statements :
Statement I : Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model.
Statement II : An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford's model. In the light of the above statements,
choose the most appropriate from the options given below.
(1) Both statement I and statement II are false
(2) Statement I is false but statement II is true
(3) Statement I is true but statement II is false
(4) Both statement I and statement II are true

2. The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the excited state of a hydrogen atom is?

3. The radius of third stationary orbit of electron for Bohr's atom is R. The radius of fourth stationary orbit will be?


Things to Remember 

  • Atoms are the basic building blocks of matter.
  • An atom consists of electrons, protons, and neutrons.
  • The first model of the atom was proposed by J. J Thomson.
  • The Impact parameter (b) is the perpendicular distance of the velocity vector from the nucleus's center.
  • The Scattering angle (θ) is the angle formed by the direction of approach of the alpha-particle and the direction of the scattered alpha-particle.
  • The spectrum of an element has a wavelength that shows definite regularities which can be classified into certain groups called spectral series.
  • The spectrum of a hydrogen atom consists of multiple series of thee line spectrum:- Pfund Series, Bracket Series, Paschen Series, Balmer Series, and Lyman Series. 
  • The postulates of Bohr’s Model of Hydrogen Atoms are given by:- The electrons of the hydrogen atom revolve around in stable orbits without emitting any radiant energy.
  • Einstein suggested that light behaves both as a material particle as well as a wave.
  • De-Broglie extended Einstein’s view and he said that all forms of matter like electrons, protons, neutrons, etc also have a dual character.

Previous Year Questions

  1. Hard X-rays for the study of fractures in bones should have a minimum wavelength of 10−11 m. The accelerating voltage for electrons in X-ray machine should be
  2. Rydberg constant is
  3. In hydrogen atom, which one of the following transitions produce a spectrum line of maximum wavelength?
  4. Band spectrum is also called​
  5. The nuclei of which one of the following pairs of nuclei are isotones?​
  6. Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be
  7. If the wavelength of 1st line of Balmer series of hydrogen is 6561 ?, the wavelength of the 2 line of series will be​
  8. What is the wavelength of the most energetic photon emitted in the Balmer series of the hydrogen atom?​
  9. Which of the following series in the spectrum of hydrogen atom lies in the visible region of the electromagnetic spectrum?​
  10. Which of the following represents isotope, isobar, isotones respectively?​
  11. If the electron in a hydrogen atom jumps from an orbit with level n1=2 to an orbit with level n2=1 the emitted radiation has a wavelength given by​
  12. In the nuclear reaction, 72X180−>[−α]Y−>[−β]Z−>[−α]A−>[−γ]P the atomic mass and atomic number of P are, respectively
  13. γ− rays are originated from
  14. As the electron in Bohr’s orbit of hydrogen atom passes from state n = 2 to, n=1, the Kinetic energy (K) and the potential energy (U) changes as
  15. What is size of gold nuclei?​
  16. How the linear velocity v of an electron in the Bohr orbit is related to its quantum number n?
  17. Sharp peak point A represents
  18. Assuming f to be the frequency of first line in Balmer series, the frequency of the immediate next (i.e. second) line is
  19. The energy of the charged particle in the nthnth level will be (h=h= Planck's constant)​
  20. If an electron is revolving around the hydrogen nucleus at a distance of 0.1nm , what should be its speed ?

Sample Questions

Ques.  An a-particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic Field B, acting normal to the direction of motion of the particles. Calculate the ratio of radii of the circular paths described by them. (3 marks)

Ans. We already know that, 

mass of proton (mp) = \(\frac{1}{4}\) x mass of alpha particle (ma)

and Charge on proton 

(qp) = \(\frac{1}{2}\) x charge of alpha particle (qa)

such that;

mv = \(\sqrt{2km}\), where k is kinetic energy.

the formula used:-

R = \(\frac{mv}{qB}\)

so, r = \(\frac{\sqrt{2km}}{qB}\)

Hence, k and b are the constant terms.

So, r ∝ \(\frac{\sqrt{m}}{q}\)

so, the relation will be:-

\(\frac{r_p}{r_a} = \sqrt{\frac{m_p}{m_a}}\frac{q_a}{q_p}\)

\(\sqrt{\frac{m_p}{4m_a}}\frac{2q_p}{q_p}\)

\(\frac{1}{1}\)

Ques. Define the distance of the closest approach. An a-particle of kinetic energy ‘K’ is bombarded on a thin gold foil. The distance of the closest approach is ‘r’. What will be the distance of the closest approach for an a-particle of double the kinetic energy?  (3 marks)

Ans. Distance of the closest approach refers to the minimum distance between the charged particles and the nucleus at which the starting kinetic energy of the particle is similar to the potential energy because of the charged nucleus.

Distance of the closest approach refers to the minimum distance between the charged particles and the nucleus at which the starting kinetic energy of the particle is similar to the potential energy because of the charged nucleus.

Ques. Write two important limitations of the Rutherford nuclear model of the atom. (2 marks)

Ans. Two important limitations are as follows:

  • The Rutherford nuclear model of the atom is not in line with Maxwell's theory and fails to give an explanation for the stability of an atom.
  • It didn't even give any explanation regarding the grouping of electrons in an atom. 

Ques. In the Rutherford scattering experiment, draw the trajectory traced by α- particles in the Coulomb field of the target nucleus and explain how this led to calculating the size of the nucleus. (5 marks)

Ans.

We can find the size of the nucleus by discovering the impact parameter b with the help of trajectories of alpha particles. The impact parameter refers to the perpendicular distance between the alpha particle having the initial velocity vector from the centerline of the nucleus when it is present at a distance from the atom.

We can find the size of the nucleus by discovering the impact parameter b with the help of trajectories of alpha particles. The impact parameter refers to the perpendicular distance between the alpha particle having the initial velocity vector from the centerline of the nucleus when it is present at a distance from the atom.

Rutherford obtained the impact parameter with the help of the following:

B = 1/4πεo . Ze2cot (θ/2) / E

Here E = KE of the α – particle.

θ = the scattering angle

Z = Atom’s atomic number

The nucleus is smaller in size than the impact parameter.

Ques.  A hydrogen atom in the ground state is excited by an electron beam of 12.5 eV energy. Find out the maximum number of lines emitted by an atom from its excited state. (3 marks)

Ans. We have: ΔE = 12.5eV

Let the electron move to n = n from n = 1 level.

ΔE = En – E1

12.5 = – (13.6/n2) – ( – 13.6/12)

12.5 = 13.6 (1 – (1/n2))

1 – (12.5/13.6) = 1/n2

1.1/13.6 = 1/n2

13.6/1.1 = n2

12.36 = n2

n = 3.5

n = 3rd

Maximum number of lines emitted by an atom from its excited state is equal to 3.

Ques. State Bohr’s quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum it belongs. (2 marks)

Ans. L = nh/2π

1/λ = R(1/42 – 1/n2)

1/λ = 1.097 * 107 m-1 (1/16 – 0)

λ = 1.46 * 10-6m

The range of infrared is between 700 nm to 1mm. Therefore it is associated with the infrared part of the spectrum.

Ques.  Calculate the orbital period of the electron in the first excited state of the hydrogen atom. (3 marks)

Ans. r = 0.53(n2/z) * 10-10m

For the former excited state n = 2

r = 0.53(22/1) * 10-10

r = 2.12 * 10-10m

V = v0 * Z/n (m/s)

V = 2.188 * 102 * Z/n (m/s)

For the initial excited state, n = 2 and Z = 1 for hydrogen atom

Therefore v = 2.188 * 102 * ½(m/s)

V = 1.094 * 102(m/s)

Orbital period = 2r/v = 2*3.14*2.12*10-10/1.094*102

Orbital period = 1.22 * 10-10 sec

Ques.  A hydrogen atom in the ground state is excited by an electron beam of 12.5 eV energy. Find out the maximum number of lines emitted by an atom from its excited state. (3 marks)

Ans. The energy present in the ground state E1 = -13.6eV

The energy that is being supplied = 12.5eV

Therefore, the energy present in the excited state refers to,

En = -13.6 + 12.5

En = -1.1eV

En = -13.6/n2

Hence, n=3

The max number of lines is equal to 3.

Ques.  State Bohr’s postulate defines stable- orbits in hydrogen atoms. How does de Broglie’s hypothesis explain the stability of these orbits? (3 marks)

Ans. Bohr’s postulate of Quantisation condition states that the electrons can only revolve in certain particular orbits that have an angular momentum of L.

Therefore the circumference of the electron that has the nth stable orbit can conserve exactly n wavelength of de-Broglie which is in association with the electron in that particular orbit.

Therefore the circumference of the electron that has the nth stable orbit can conserve exactly n wavelength of de-Broglie which is in association with the electron in that particular orbit.

Ques.  Find out the wavelength of the electron orbiting in the ground state of the hydrogen atom. (2 marks)

Ans. The radius of the hydrogen atom’s ground state is equal to 0.53Ao = 0.53 * 10-10m

Obtaining from the de Broglie relation 2πr = nλ

For the ground state n = 1

2 * 3 .14 * 0.53 * 10-10 = 1 * λ

Hence, λ = 3.32 * 10-10m

= 3.32 Ao

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates


Check-Out: 

CBSE CLASS XII Related Questions

  • 1.
    Answer the following giving reason:
    (a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
    (b) The saturation current in case (a) is different for different intensity.
    (c) If one goes on increasing the wavelength of light incident on a metal sur face, keeping its intensity constant, emission of photoelectrons stops at a certain wavelength for this metal.


      • 2.
        In the circuit, three ideal cells of e.m.f. \( V \), \( V \), and \( 2V \) are connected to a resistor of resistance \( R \), a capacitor of capacitance \( C \), and another resistor of resistance \( 2R \) as shown in the figure. In the steady state, find (i) the potential difference between P and Q, (ii) the potential difference across capacitor C.
        potential difference across capacitor C


          • 3.
            A charge \( -6 \mu C \) is placed at the center B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge \( +5 \mu C \) is moved from point ‘C’ to point ‘A’ along the circumference. Calculate the work done on the charge.
            work done on the charge


              • 4.
                Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).


                  • 5.
                    A rectangular glass slab ABCD (refractive index 1.5) is surrounded by a transparent liquid (refractive index 1.25) as shown in the figure. A ray of light is incident on face AB at an angle \(i\) such that it is refracted out grazing the face AD. Find the value of angle \(i\).
                    A rectangular glass slab ABCD (refractive index 1.5)


                      • 6.
                        The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

                          • \( \frac{\alpha}{r} \)
                          • \( \frac{\alpha r^2}{2} \)
                          • \( \frac{\alpha}{2r^2} \)
                          • \( -\frac{\alpha}{r} \)
                        CBSE CLASS XII Previous Year Papers

                        Comments


                        No Comments To Show