Heisenberg Uncertainty Principle-Derivation, Detailed Explanation, Formula

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Heisenberg’s uncertainty principle states that it is impossible to measure or calculate exactly both the position and the momentum of an object. This principle is based on the wave-particle duality of matter.

Although Heisenberg’s uncertainty principle can be ignored in the macroscopic world (the uncertainties in the position and velocity of objects with relatively large masses are negligible), it holds significant value in the quantum world.

Since atoms and subatomic particles have very small masses, any increase in the accuracy of their positions will be accompanied by an increase in the uncertainty associated with their velocities.

In the field of quantum mechanics, Heisenberg’s uncertainty principle is a fundamental theory that explains why it is impossible to measure more than one quantum variable simultaneously.

Another implication of the uncertainty principle is that it is impossible to accurately measure the energy of a system in some finite amount of time. 

Why Is It Impossible to Measure Both Position and Momentum Simultaneously?
In order to illustrate Heisenberg’s uncertainty principle, consider an example where the position of an electron is measured.
In order to measure the position of an object, a photon must collide with it and return to the measuring device.
Since photons hold some finite momentum, a transfer of momenta will occur when the photon collides with the electron. This transfer of momenta will cause the momentum of the electron to increase.
Thus, any attempt at measuring the position of a particle will increase the uncertainty in the value of its momentum.
Applying the same example to a macroscopic object (say a basketball), it can be understood that Heisenberg’s uncertainty principle has a negligible impact on measurements in the macroscopic world.
While measuring the position of a basketball, there will still be a transfer of momentum from the photons to the ball.
However, the mass of the photon is much smaller than the mass of the ball. Therefore, any momentum imparted by the photon to the ball can be neglected.  

Heisenberg’s uncertainty principle imposes a restriction on the accuracy of simultaneous measurement of position and momentum. The more precise our measurement of position is, the less accurate will be our momentum measurement and vice-versa. The physical origin of Heisenberg’s uncertainty principle is with the quantum system. Determination of position by performing a measurement on the system disturbs it sufficiently to make the determination of q imprecise and vice-versa. We will learn about the principle in detail below.

What Is Heisenberg’s Uncertainty Principle?

Heisenberg’s uncertainty principle states that for particles exhibiting both particle and wave nature, it will not be possible to accurately determine both the position and velocity at the same time.

The principle is named after  German physicist Werner Heisenberg, who proposed the uncertainty principle in the year 1927. This principle was formulated when Heisenberg was trying to build an intuitive model of quantum physics. He discovered that there were certain fundamental factors that limited our actions in knowing certain quantities.

This principle basically highlights that simultaneous measurement of position and the velocity or momentum of microscopic matter waves will have an error such that the product of the error in measurement of position and momentum is equal or more than an integral multiple of a constant.

Origin of the Uncertainty Principle

One of the crucial points for the genesis of the uncertainty principle is solely due to the dual nature of a wave-particle.

Every particle is said to have a wave nature, and the probability of finding particles is maximum, where the waveforms are the greatest. If the particle has greater undulation, the wavelength becomes more indistinct or vague.

However, we are able to determine the momentum of the particle. From what we have learnt so far, we can say that the particles that have definite positions will have no fixed velocity.

On the other hand, a particle with a well-defined wavelength will show a definitive or precise velocity. All in all, if we get an accurate reading of one quantity, it will only lead to large uncertainty in the measurement of the other.

Heisenberg Uncertainty Principle Formula and Application

If ∆x is the error in position measurement and ∆p is the error in the measurement of momentum, then

\(\begin{array}{l}\Delta X \times \Delta p \ge \frac{h}{4\pi }\end{array} \)

Since momentum, p = mv, Heisenberg’s uncertainty principle formula can be alternatively written as,

\(\begin{array}{l}\Delta X \times \Delta mv \ge \frac{h}{4\pi }\end{array} \)

or     

\(\begin{array}{l}\Delta X \times \Delta m \times \Delta v \ge\frac{h}{4\pi }\end{array} \)

Where, ∆V is the error in the measurement of velocity, and assuming mass remains constant during the experiment,

\(\begin{array}{l}\Delta X \times \Delta V \ge\frac{h}{4\pi m}\end{array} \)

.

Accurate measurement of position or momentum automatically indicates larger uncertainty (error) in the measurement of the other quantity.

Applying the Heisenberg principle to an electron in an orbit of an atom, with h = 6.626 ×10-34Js and m= 9.11 ×10-31Kg,

\(\begin{array}{l}\Delta X \times \Delta V \ge\frac{6.626\times {{10}^{-34}}}{4\times 3.14\times 9.11\times {{10}^{-31}}}=10^{-4}\ m^2\ s^{-1}\end{array} \)

If the position of the electron is measured accurately to its size (10-10m), then the error in the measurement of its velocity will be equal to or larger than 106m or 1000 km.

Heisenberg’s principle applies to only dual-natured microscopic particles and not to macroscopic particles whose wave nature is minimal.

Explaining Heisenberg Uncertainty Principle with an Example

Electromagnetic radiations and microscopic matter waves exhibit a dual nature of mass/momentum and wave character. The position and velocity/momentum of macroscopic matter waves can be determined accurately and simultaneously. For example, the location and speed of a moving car can be determined simultaneously with minimum error. But, in microscopic particles, it will not be possible to fix the position and measure the velocity/momentum of the particle simultaneously.

An electron in an atom has a mass of 9.91 × 10-31 kg. Naked eyes cannot see such small particles. A powerful light may collide with the electron and illuminate it. Illumination helps in identifying and measuring the position of the electron. The collision of the powerful light source, while helping in identification, increases the momentum of the electron and makes it move away from the initial position.

Thus, when fixing the position, the velocity/momentum of the particle would have changed from the original value. Hence, when the position is exact, the error occurs in measuring velocity or momentum. In the same way, the measurement of momentum, accurately, will change the position.

Hence, at any point in time, either position or momentum can only be measured accurately.

Simultaneous measurement of both of them will have an error in both position and momentum. Heisenberg quantified the error in the measurement of both position and momentum simultaneously.

Heisenberg’s γ-ray Microscope

A striking thought experiment illustrating the uncertainty principle is Bohr’s/Heisenberg’s Gamma-ray microscope.

To observe a particle, say an electron, we shine it with the light ray of wavelength λ and collect the Compton scattered light in a microscope objective whose diameter subtends an angle θ with the electron, as shown in the figure below

Heisenberg’s uncertainty principle states that it is impossible to measure or calculate exactly both the position and the momentum of an object. This principle is based on the wave-particle duality of matter.

The precision with which the electron can be located, Delta x, is defined by the resolving power of the microscope,

\(\begin{array}{l}sin \theta =\frac{\lambda }{\Delta x}\Rightarrow \Delta x=\frac{\lambda }{sin \theta }\end{array} \)

It appears that by making λ small, we choose γ-ray, and by making sin θ large, Delta x can be made as small as desired. But, according to the uncertainty principle, we can do so only at the expense of our knowledge of the x-component of electron momentum.

In order to record the Compton scattered photon by the microscope, the photon must stay in the cone of angle θ, and hence its x-component of the momentum can vary within ±(h/λ) sin θ. This implies that the magnitude of the recoil momentum of the electron is uncertain by

\(\begin{array}{l}\Delta p_{x}=\frac{2h}{\lambda }sin \theta\end{array} \)

The product of the uncertainty yields,

\(\begin{array}{l}\Delta x\Delta p_{x}=\frac{\lambda }{sin \theta }\frac{2h}{\lambda }sin \theta =4\pi h\end{array} \)

Is Heisenberg’s Uncertainty Principle Noticeable in All Matter Waves?

Heisenberg’s principle is applicable to all matter waves. The measurement error of any two conjugate properties, whose dimensions happen to be joule sec, like position-momentum, the time-energy will be guided by Heisenberg’s value.

But, it will be noticeable and of significance only for small particles like an electron with very low mass. A bigger particle with a heavy mass will show the error to be very small and negligible.

Heisenberg Uncertainty Principle Equations

Heisenberg’s uncertainty principle can be considered as a very precise mathematical statement that describes the nature of quantum systems. As such, we often consider two common equations related to the uncertainty principle, and they are as follows:

Equation 1: ∆X ⋅ ∆p ~ ħ

Equation 2:  ∆E ⋅ ∆t ~ ħ

Where,

ħ = value of Planck’s constant divided by 2*pi
∆X = uncertainty in the position
∆p = uncertainty in momentum
∆E = uncertainty in the energy
∆t = uncertainty in time measurement

Solved Numerical Problems on Heisenberg’s Uncertainty Principle

1. If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h / 4pm × 0.05 nm. Is there any problem in defining this value?

a) ∆x = 2×10-12m;

\(\begin{array}{l}\Delta X \times \Delta mV \ge\frac{h}{4\pi } = \frac{6.626\times {{10}^{-34}}}{4\times 3.14}\end{array} \)

\(\begin{array}{l}\because \Delta mV \ge \frac{h}{4\pi \Delta x} \ge \frac{6.626\times {{10}^{-34}}}{4\times 3.14\times 2\times {{10}^{-12}}}\,\end{array} \)

= 2.64 × 10-23 Kg m s-1

b) Momentum

\(\begin{array}{l}mv = \frac{h\times 5\times {{10}^{-11}}}{4\times {{10}^{-12}}}=\frac{6.626\times {{10}^{-34}}\times 5\times {{10}^{-11}}}{4\times {{10}^{-12}}}\,\,\end{array} \)

= 28 × 10-33

Error in momentum measurement is 1010 times larger than the actual momentum. The given momentum will not be acceptable.

2. Position of a chloride ion on a material can be determined to a maximum error of 1μm. If the mass of the chloride ion is 5.86 × 10-26kg, what will be the error in its velocity measurement?

∆x =  10-6 m;

\(\begin{array}{l}\Delta X \times \Delta mV \ge\frac{h}{4\pi }= \frac{6.626\times {{10}^{-34}}}{4\times 3.14}\end{array} \)

   = 5.28×10-35Js

\(\begin{array}{l}\because \Delta V \ge \frac{h}{4\pi m\Delta x}\ge \frac{6.626\times {{10}^{-34}}}{4\times 3.14\times 5.86\times {{10}^{-26}}\times {{10}^{-6}}}\,\end{array} \)

= 9  × 10-4m s-1

3. The lifetime of an excited state of an atom is 3 × 10-3s. What is the minimum uncertainty in its energy in eV?

Time and energy are conjugate pairs with the Js unit. The product of measurement error is given by Heisenberg’s principle.

\(\begin{array}{l}\Delta t \times \Delta E \ge \frac{h}{4\pi } = \frac{6.626\times {{10}^{-34}}}{4\times 3.14}\end{array} \)

= 5.28×10-35Js

Assuming a maximum error in the measurement of lifetime equal to that of lifetime = 3 ×10-3s

\(\begin{array}{l}\Delta E \ge \frac{h}{4\pi m\Delta x}=\frac{1}{3\times {{10}^{-3}}} \times 5.28 \times 10^{-35}\ \text{J}\end{array} \)

⸪ 1 Joule = 6.242 × 1018ev,

Uncertainty in the determination of energy of the atom = ∆E

\(\begin{array}{l}=6.22\times 10^{18}\times \frac{1}{3\times {{10}^{-3}}}\times 5.28\times 10^{-35}\end{array} \)

= 1.1 × 10-13

4. A wet ball weighing 10.1gm has a water of 0.1 g on it. The ball is moving with a constant velocity with an uncertainty of momentum of 10-6 kg m/s. What will be the uncertainty in the measurement of the position of the ball, water and electron in the water molecule?

\(\begin{array}{l}\Delta X \times \Delta p \ge \frac{h}{4\pi }\end{array} \)

Velocity being constant, uncertainty in the measurement of the momentum is associated with the mass of the matter.

Uncertainty in the momentum of the dry ball = mass ×10-6   = 10×10-3×10-6 Kg m s-1.

Uncertainty in the momentum of the water    = mass ×10-6 = 0.1×10-3×10-6  Kg m s-1.

Uncertainty in the momentum of the electron = mass ×10-6 = 9×10-31×10-6  Kg m s-1.

Uncertainty in position measurement is inversely proportional to the uncertainty in momentum.

\(\begin{array}{l}\Delta X \ge \frac{h}{4\pi m\Delta p} \propto \frac{1}{\Delta p}\end{array} \)

∆Xb :  ∆Xw : ∆Xe

\(\begin{array}{l} = \frac{1}{{{10}^{-8}}}\,:\,\,\frac{1}{{{10}^{-10}}}:\,\frac{1}{9\times {{10}^{-37}}}\,\end{array} \)

=    108 :    1010   :    1.1 × 1036   or

=  1      :   102     :    1028

5. Determine the minimum uncertainties in the positions of the electron if their speeds are known with a precision of 3.0 × 10−3m/s.

Solution:

Δu = 3.0×10−3m/s

Uncertained momentum Δp = m Δu

Uncertainty in position Δx = ℏ/(2Δp)

For electron

Δp = m Δu

= (9.1×10−31kg × 3.0×10−3)

Δp  = 2.73 × 10-33 kg.m/s

\(\begin{array}{l}\Delta x = \frac{h}{\Delta p }\end{array} \)

Δx = 0.12 m

Frequently‌ ‌Asked‌ ‌Questions‌ ‌on‌ Heisenberg’s Uncertainty Principle

Q1

Who was the first to come up with the idea of finding an electron in an orbital?

Heisenberg was the first one to think of or develop this idea.

Q2

Is it possible to determine that the electron is located at a certain exact location with 100% accuracy at a fixed time through measurement?

Q3

Which scientists proposed the idea of the uncertainty principle and the concept of the wave nature of matter?

It was Heisenberg and de Broglie.

Q4

What violates the Heisenberg uncertainty principle?

If an object travelling through spacetime can loop back in time in a certain way, then its trajectory can allow a pair of its components to be measured with perfect accuracy, violating Heisenberg’s uncertainty principle.

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.

JEE Syllabus for Chapter - ATOMIC STRUCTURE

According to the JEE syllabus, the "Atomic Structure" chapter covers key concepts like: Nature of electromagnetic radiation, photoelectric effect, spectrum of the hydrogen atom, Bohr model of ahydrogen atom - its postulates, derivation of the relations for the energy of the electron and radii of the different orbits, limitations of Bohr's model, dual nature of matter, de Broglie's relationship, Heisenberguncertainty principle, elementary ideas of quantum mechanics, the quantum mechanical model of the atomand its important features, concept of atomic orbitals as one-electron wave functions, variation of Ψ and Ψ2 with r for 1s and 2s orbitals, various quantum numbers (principal, angular momentumand magneticquantum numbers) and their significance, shapes of s, p and d - orbitals, electron spin and spin quantumnumber, rules for filling electrons in orbitals – Aufbau principle, Pauli's exclusion principle and Hund'srule, electronic configuration of elements and extra stability of half-filled and completely filled orbitals.

NEET Syllabus for Chapter - ATOMIC STRUCTURE

According to the NEET syllabus, the "Atomic Structure" chapter covers key concepts like: subatomic particles (protons, electrons, neutrons), atomic number and mass number, various atomic models (Dalton's, Thomson's, Rutherford's, Bohr's), quantum mechanical model (Schrödinger's equation, quantum numbers - principal, azimuthal, magnetic, and spin), shapes of orbitals (s, p, d), electronic configuration of elements based on Aufbau principle, Pauli exclusion principle, and Hund's rule; including the limitations of Bohr's model and the concept of dual nature of matter with de Broglie's relationship and Heisenberg's uncertainty principle. 

MCQs on Structure of Atom for Class 11 CBSE Board Exam

Here are some multiple-choice questions (MCQs) on Structure of Atom for Class 11 CBSE along with detailed explanations:


1. Which of the following statements about the nucleus of an atom is correct?

A) It contains protons and neutrons
B) It has a negative charge
C) It occupies most of the volume of the atom
D) It is responsible for chemical properties of an atom

Answer: A) It contains protons and neutrons

Explanation:
The nucleus of an atom contains protons (positively charged) and neutrons (neutral). The electrons revolve around the nucleus in different energy levels. The nucleus has a positive charge due to the presence of protons. It occupies a very small volume but contributes to almost the entire mass of the atom.


2. The total number of electrons that can be accommodated in the second shell (L-shell) is:

A) 2
B) 8
C) 18
D) 32

Answer: B) 8

Explanation:
The maximum number of electrons in a shell is given by 2n², where n is the shell number.
For the second shell (n = 2):
Max electrons = 2(2²) = 8.


3. Which of the following is NOT a postulate of Bohr’s atomic model?

A) Electrons revolve around the nucleus in fixed orbits
B) Electrons emit or absorb energy when they jump between orbits
C) Energy levels are quantized
D) Electrons can have any random energy value

Answer: D) Electrons can have any random energy value

Explanation:
Bohr's model states that electrons revolve in fixed energy levels, and they cannot have arbitrary energy. Electrons only gain or lose energy when they transition between these discrete orbits.


4. The isotope of hydrogen that contains one proton and two neutrons is:

A) Protium
B) Deuterium
C) Tritium
D) None of these

Answer: C) Tritium

Explanation:

  • Protium (¹H) → 1 proton, 0 neutrons
  • Deuterium (²H) → 1 proton, 1 neutron
  • Tritium (³H) → 1 proton, 2 neutrons

5. The wave nature of electrons was proposed by:

A) Bohr
B) Heisenberg
C) de Broglie
D) Rutherford

Answer: C) de Broglie

Explanation:
Louis de Broglie proposed that electrons exhibit both particle and wave nature (wave-particle duality). His equation λ = h/mv relates the wavelength (λ) of a moving particle to its momentum.


6. The uncertainty principle was proposed by:

A) Bohr
B) Heisenberg
C) Rutherford
D) Schrodinger

Answer: B) Heisenberg

Explanation:
Heisenberg's Uncertainty Principle states that it is impossible to simultaneously determine the exact position and momentum of an electron.


7. The quantum number that describes the shape of an orbital is:

A) Principal quantum number (n)
B) Azimuthal quantum number (l)
C) Magnetic quantum number (m)
D) Spin quantum number (s)

Answer: B) Azimuthal quantum number (l)

Explanation:
The Azimuthal quantum number (l) determines the shape of orbitals:

  • s-orbital (l = 0) → Spherical
  • p-orbital (l = 1) → Dumbbell
  • d-orbital (l = 2) → Complex
  • f-orbital (l = 3) → More complex

8. Which of the following orbitals cannot exist?

A) 1s
B) 2p
C) 3f
D) 4d

Answer: C) 3f

Explanation:
For an orbital to exist, the Azimuthal quantum number (l) must satisfy:
l = 0 to (n-1), where n is the principal quantum number.
For n = 3, possible l values: 0 (s), 1 (p), 2 (d) → No f-orbital.


9. Which of the following elements has the electronic configuration: 1s² 2s² 2p⁶ 3s¹?

A) Sodium (Na)
B) Magnesium (Mg)
C) Aluminium (Al)
D) Potassium (K)

Answer: A) Sodium (Na)

Explanation:

  • 1s² 2s² 2p⁶ 3s¹ is the electronic configuration of sodium (Na) (Atomic number 11).
  • Magnesium (Mg) = 1s² 2s² 2p⁶ 3s²
  • Aluminium (Al) = 1s² 2s² 2p⁶ 3s² 3p¹

10. The shape of an s-orbital is:

A) Spherical
B) Dumbbell
C) Double dumbbell
D) Complex

Answer: A) Spherical

Explanation:
The s-orbital is spherically symmetric around the nucleus. p-orbitals are dumbbell-shaped.


11. The number of orbitals present in the third energy level (n = 3) is:

A) 3
B) 9
C) 18
D) 5

Answer: B) 9

Explanation:
Total orbitals in an energy level =
For n = 3, orbitals = 3² = 9
(1 s-orbital, 3 p-orbitals, 5 d-orbitals)


12. If an electron has quantum numbers n = 3, l = 2, what type of orbital is it in?

A) 3s
B) 3p
C) 3d
D) 3f

Answer: C) 3d

Explanation:

  • n = 3 (Third shell)
  • l = 2 corresponds to d-orbital3d orbital

13. The maximum number of electrons that can be accommodated in an orbital is:

A) 1
B) 2
C) 4
D) 6

Answer: B) 2

Explanation:
Each orbital can hold a maximum of 2 electrons with opposite spins, as per Pauli's exclusion principle.


14. Which quantum number determines the energy of an electron in a hydrogen atom?

A) Principal quantum number (n)
B) Azimuthal quantum number (l)
C) Magnetic quantum number (m)
D) Spin quantum number (s)

Answer: A) Principal quantum number (n)

Explanation:
For hydrogen-like atoms, the energy of an electron depends only on n.


15. The concept of orbitals was introduced by:

A) Bohr
B) Rutherford
C) Schrodinger
D) Heisenberg

Answer: C) Schrodinger

Explanation:
Schrodinger’s wave equation introduced orbitals (regions of high probability of finding an electron), replacing Bohr’s fixed orbits.

Assertion and Reason (A-R) type questions on Structure of Atom for Class 11 CBSE Board Exam

Here are some Assertion and Reason (A-R) type questions on Structure of Atom for Class 11 CBSE, along with detailed explanations.


How to Answer Assertion-Reason Questions?

Each question consists of two statements:

  • Assertion (A): A statement of fact.
  • Reason (R): An explanation for the assertion.

You must choose the correct option:

  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is NOT the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.

1. Assertion (A): The nucleus of an atom contains protons and neutrons.

Reason (R): The electrons in an atom revolve around the nucleus in fixed orbits.

Answer: Option (2) (Both A and R are true, but R is not the correct explanation of A.)

🔹 Explanation:

  • The nucleus is composed of protons and neutrons, which was confirmed by Rutherford’s experiment.
  • Electrons revolve around the nucleus in fixed orbits, as described in Bohr’s atomic model.
  • However, the presence of electrons in orbits does not explain why the nucleus contains protons and neutrons.

2. Assertion (A): The maximum number of electrons in a shell is given by 2n².

Reason (R): Electrons in an atom are arranged in different shells around the nucleus.

Answer: Option (1) (Both A and R are true, and R is the correct explanation of A.)

🔹 Explanation:

  • The Bohr-Bury rule states that the maximum number of electrons in a shell = 2n², where n is the shell number.
  • Electrons are distributed in different shells, and this follows the energy level distribution principle.

Example:

  • K shell (n = 1): 2(1²) = 2 electrons
  • L shell (n = 2): 2(2²) = 8 electrons
  • M shell (n = 3): 2(3²) = 18 electrons

Thus, R correctly explains A.


3. Assertion (A): The mass of an atom is concentrated in the nucleus.

Reason (R): The electrons have negligible mass compared to protons and neutrons.

Answer: Option (1) (Both A and R are true, and R is the correct explanation of A.)

🔹 Explanation:

  • The nucleus contains protons and neutrons, which are heavy particles.
  • Electrons are much lighter (mass = 1/1836 of a proton), so their contribution to atomic mass is negligible.
  • Thus, the nucleus contains almost all the mass of an atom, and R explains A correctly.

4. Assertion (A): The energy of electrons in an atom is quantized.

Reason (R): Electrons can exist at any random energy level.

Answer: Option (3) (A is true, but R is false.)

🔹 Explanation:

  • According to Bohr’s model, electrons can only occupy specific, discrete energy levels (quantized energy states).
  • Electrons cannot have arbitrary energy values.
  • Hence, A is true, but R is false because electrons follow quantized energy levels.

5. Assertion (A): The Heisenberg Uncertainty Principle states that the exact position and momentum of an electron cannot be simultaneously determined.

Reason (R): Electrons move in fixed circular orbits around the nucleus.

Answer: Option (3) (A is true, but R is false.)

🔹 Explanation:

  • Heisenberg’s Uncertainty Principle states:
    Δx × Δp ≥ h/4π
    (where Δx = uncertainty in position, Δp = uncertainty in momentum)
  • This means we cannot know both the exact position and momentum of an electron at the same time.
  • Bohr’s model (fixed orbits) was later replaced by Schrodinger’s model (probability orbitals).
  • R is false because electrons do not move in fixed orbits but in probabilistic regions (orbitals).

6. Assertion (A): The azimuthal quantum number (l) determines the shape of an orbital.

Reason (R): The principal quantum number (n) determines the size of the orbital.

Answer: Option (1) (Both A and R are true, and R is the correct explanation of A.)

🔹 Explanation:

  • The Azimuthal quantum number (l) determines the shape of an orbital (s, p, d, f).
  • The Principal quantum number (n) determines the size and energy of an orbital.
  • Since both statements are correct and R correctly explains A, the answer is Option (1).

7. Assertion (A): The 3f orbital exists.

Reason (R): The azimuthal quantum number (l) for an f-orbital is 3.

Answer: Option (4) (A is false, but R is true.)

🔹 Explanation:

  • l = 3 represents an f-orbital, but for an orbital to exist, n must be greater than l.
  • For n = 3, the possible values of l are 0 (s), 1 (p), and 2 (d).
  • f-orbital (l = 3) is possible only when n ≥ 4, so 3f does not exist.
  • Hence, A is false, but R is true.

8. Assertion (A): The Pauli Exclusion Principle states that no two electrons in an atom can have the same set of four quantum numbers.

Reason (R): An orbital can accommodate a maximum of two electrons with opposite spins.

Answer: Option (1) (Both A and R are true, and R is the correct explanation of A.)

🔹 Explanation:

  • Pauli's Exclusion Principle states that no two electrons in an atom can have identical quantum numbers.
  • This is because each orbital can hold only two electrons with opposite spins.
  • Since R explains A correctly, the answer is Option (1).

9. Assertion (A): The probability of finding an electron is maximum near the nucleus.

Reason (R): The probability density of an electron is given by the wave function (ψ²).

Answer: Option (1) (Both A and R are true, and R is the correct explanation of A.)

🔹 Explanation:

  • The Schrodinger wave equation gives ψ², which represents the probability density of finding an electron in a region.
  • For s-orbitals, the highest probability is near the nucleus.
  • Since R explains A correctly, the answer is Option (1).

FAQs (Important Questions & Answers) on Structure of Atom – Class 11

1. What is the structure of an atom?

The structure of an atom consists of a nucleus containing protons and neutrons, surrounded by electrons moving in discrete energy levels or shells around the nucleus.

2. Who discovered the atom?

The concept of the atom was first proposed by John Dalton in his Atomic Theory (1808). Later, significant contributions were made by J.J. Thomson, Rutherford, and Bohr to explain its structure.

3. What are the fundamental particles of an atom?

The three fundamental particles of an atom are:

  • Proton (p⁺) – Positively charged, found in the nucleus.
  • Neutron (n⁰) – Neutral, found in the nucleus.
  • Electron (e⁻) – Negatively charged, revolves around the nucleus.

4. What is Dalton’s Atomic Theory?

Dalton’s Atomic Theory (1808) states that:

  • Matter is made of indivisible atoms.
  • Atoms of an element are identical in mass and properties.
  • Atoms combine in whole-number ratios to form compounds.
  • Atoms cannot be created or destroyed in a chemical reaction.

5. What was J.J. Thomson’s Model of the Atom?

J.J. Thomson proposed the Plum Pudding Model (1897), where an atom was visualized as a positively charged sphere with negatively charged electrons embedded in it, like plums in a pudding. However, this model was later disproven by Rutherford’s experiment.

6. What is Rutherford’s Atomic Model?

Rutherford’s Gold Foil Experiment (1911) led to the discovery that:

  • An atom has a dense positively charged nucleus.
  • Electrons revolve around the nucleus.
  • Most of the atom is empty space.
    This model could not explain atomic stability, which was later addressed by Bohr’s model.

7. What is Bohr’s Atomic Model?

Niels Bohr (1913) proposed that:

  • Electrons revolve in fixed energy levels (shells).
  • They do not lose energy while in stable orbits.
  • Energy is absorbed or emitted when an electron jumps between energy levels.

8. What are Quantum Numbers?

Quantum numbers define the position and energy of an electron in an atom:

  1. Principal Quantum Number (n) – Represents the main energy level.
  2. Azimuthal Quantum Number (l) – Defines the shape of the orbital.
  3. Magnetic Quantum Number (mₗ) – Indicates the orientation of the orbital.
  4. Spin Quantum Number (mₛ) – Describes the spin of an electron (+½ or -½).

9. What is the Heisenberg Uncertainty Principle?

Proposed by Werner Heisenberg, it states that it is impossible to simultaneously determine the exact position and momentum of an electron in an atom.

10. What is the difference between an orbit and an orbital?

  • Orbit – A fixed path in which electrons revolve (Bohr’s Model).
  • Orbital – A 3D region in space where the probability of finding an electron is highest (Quantum Mechanical Model).

11. What is the Aufbau Principle?

The Aufbau Principle states that electrons fill atomic orbitals in order of increasing energy levels, i.e., lower-energy orbitals are filled first before higher ones.

12. What is Hund’s Rule?

Hund’s Rule states that in degenerate (equal energy) orbitals, electrons fill each orbital singly before pairing up.

13. What is Pauli’s Exclusion Principle?

Pauli’s Exclusion Principle states that no two electrons in an atom can have the same set of four quantum numbers. This means an atomic orbital can hold a maximum of two electrons with opposite spins.

14. What is the Quantum Mechanical Model of the Atom?

Developed by Schrödinger, this model describes electrons as wave-like particles with a probability distribution around the nucleus instead of fixed orbits. It introduces atomic orbitals (s, p, d, f) as probable electron locations.

15. What are the types of orbitals in an atom?

There are four types of orbitals:

  • s-orbital – Spherical shape (holds max 2 electrons).
  • p-orbital – Dumbbell shape (holds max 6 electrons).
  • d-orbital – Complex shape (holds max 10 electrons).
  • f-orbital – Complex shape (holds max 14 electrons).

16. What is the significance of the atomic number and mass number?

  • Atomic Number (Z) – Number of protons in an atom.
  • Mass Number (A) – Sum of protons and neutrons in an atom’s nucleus.

17. What are Isotopes, Isobars, and Isotones?

  • Isotopes – Same atomic number, different mass number (e.g., Hydrogen: ¹H, ²H, ³H).
  • Isobars – Same mass number, different atomic number (e.g., ¹⁴C and ¹⁴N).
  • Isotones – Same number of neutrons, different atomic and mass numbers (e.g., ¹⁴C and ¹⁵N).

18. What is the Dual Nature of Electrons?

Proposed by de Broglie, it states that electrons exhibit both particle and wave-like properties, known as wave-particle duality.

19. What is the Electronic Configuration of an Atom?

The electronic configuration of an atom describes how electrons are distributed in different orbitals. It follows the Aufbau principle, Hund’s rule, and Pauli’s exclusion principle (e.g., Oxygen (O) = 1s² 2s² 2p⁴).

20. Why is the Bohr model still used despite its limitations?

Although the Bohr model fails for multi-electron atoms and does not explain fine spectral lines, it is still useful because it provides a simple and understandable representation of electron energy levels.

📚 CBSE Class 11 Chemistry: Structure of Atom – Complete Syllabus Overview

Are you a Class 11 CBSE student aiming to master "Structure of Atom"? Here’s a comprehensive breakdown of the chapter to help you focus your studies!

Case Study-Based MCQs on Structure of Atom – Class 11 CBSE

Case study-based questions are designed to test your analytical skills and conceptual understanding. Below are five case studies with multiple-choice questions (MCQs) and detailed explanations.


Case Study 1: Discovery of Atomic Structure

The structure of an atom has been explored through various experiments. J.J. Thomson discovered electrons using a cathode ray tube experiment, while Goldstein discovered protons using canal rays. Rutherford’s alpha particle scattering experiment led to the discovery of a dense nucleus at the center of the atom. Later, Bohr’s model refined our understanding by proposing that electrons move in discrete energy levels.

1.1 What was the key observation in Rutherford’s experiment?

A) Most alpha particles were deflected back
B) Alpha particles passed through the gold foil without deflection
C) All alpha particles got absorbed by the gold foil
D) Electrons were ejected from the gold foil

Answer: B) Alpha particles passed through the gold foil without deflection

🔹 Explanation: Rutherford’s experiment showed that most of the space in an atom is empty, as most alpha particles passed straight through the foil. However, a few were deflected, indicating the presence of a dense, positively charged nucleus.


1.2 What did J.J. Thomson’s experiment conclude?

A) Atoms have a nucleus
B) Atoms are indivisible
C) Atoms contain negatively charged particles
D) Atoms are mostly empty space

Answer: C) Atoms contain negatively charged particles

🔹 Explanation: J.J. Thomson’s cathode ray experiment showed the presence of negatively charged electrons, leading to the "plum pudding model" of the atom.


Case Study 2: Bohr’s Atomic Model

Niels Bohr proposed that electrons move in fixed orbits (energy levels) around the nucleus without losing energy. He introduced the quantization of energy levels and explained the emission spectra of hydrogen.

2.1 According to Bohr’s model, what happens when an electron jumps from a higher to a lower energy level?

A) The atom becomes unstable
B) The electron absorbs energy
C) The electron loses energy in the form of radiation
D) The electron disappears

Answer: C) The electron loses energy in the form of radiation

🔹 Explanation: When an electron moves from a higher energy level to a lower one, it releases energy in the form of light (photon), which creates the atomic emission spectrum.


2.2 Which of the following is NOT a postulate of Bohr’s model?

A) Electrons revolve around the nucleus in circular orbits
B) Electrons can have any energy within an orbit
C) Energy levels are quantized
D) Electrons do not lose energy while revolving

Answer: B) Electrons can have any energy within an orbit

🔹 Explanation: According to Bohr, electrons can only exist in specific, quantized energy levels, meaning they cannot have any arbitrary energy.


Case Study 3: Quantum Mechanical Model

The quantum mechanical model, developed by Schrodinger, replaced Bohr’s model. It introduced the concept of orbitals, where electrons are found as a probability distribution rather than fixed orbits. The model is based on wave-particle duality and Heisenberg’s Uncertainty Principle.

3.1 What does Heisenberg’s Uncertainty Principle state?

A) Electrons are present in fixed orbits
B) We cannot determine both position and momentum of an electron simultaneously
C) Electrons revolve around the nucleus like planets
D) Electrons follow a predictable path

Answer: B) We cannot determine both position and momentum of an electron simultaneously

🔹 Explanation: Heisenberg’s Uncertainty Principle states that it is impossible to simultaneously know the exact position and momentum of an electron, leading to the concept of probability orbitals.


3.2 Which scientist introduced the wave equation to describe electron behavior?

A) Bohr
B) Heisenberg
C) Schrodinger
D) Rutherford

Answer: C) Schrodinger

🔹 Explanation: Erwin Schrodinger developed the wave equation, which describes the probability of finding an electron in a given region of space, known as an orbital.


Case Study 4: Quantum Numbers

Each electron in an atom is described by four quantum numbers:

  1. Principal quantum number (n) → Determines the energy level
  2. Azimuthal quantum number (l) → Determines the shape of orbitals
  3. Magnetic quantum number (m) → Determines the orientation of orbitals
  4. Spin quantum number (s) → Determines electron spin

4.1 What does the principal quantum number (n) determine?

A) Shape of the orbital
B) Orientation of the orbital
C) Size and energy of the orbital
D) Spin of the electron

Answer: C) Size and energy of the orbital

🔹 Explanation: The principal quantum number (n) determines the size of the electron cloud and the energy level where the electron resides.


4.2 How many orbitals are present in the third energy level (n = 3)?

A) 3
B) 9
C) 18
D) 5

Answer: B) 9

🔹 Explanation: The number of orbitals in an energy level is given by .
For n = 33² = 9 orbitals (1s, 3p, 5d).


Case Study 5: Electronic Configuration and Periodicity

The electronic configuration of an atom follows the Aufbau principle, Pauli Exclusion Principle, and Hund’s Rule. Elements in the periodic table are arranged based on their atomic number and valence electron configuration.

5.1 Which rule states that electrons fill orbitals in order of increasing energy?

A) Hund’s Rule
B) Aufbau Principle
C) Pauli’s Exclusion Principle
D) Heisenberg’s Principle

Answer: B) Aufbau Principle

🔹 Explanation: The Aufbau Principle states that electrons occupy the lowest energy orbitals first before filling higher ones.


5.2 The electronic configuration of an element is 1s² 2s² 2p⁶ 3s². What is the element?

A) Magnesium (Mg)
B) Sodium (Na)
C) Aluminium (Al)
D) Oxygen (O)

Answer: A) Magnesium (Mg)

🔹 Explanation:

  • 1s² 2s² 2p⁶ 3s² corresponds to an atomic number of 12, which is magnesium (Mg).

CBSE Board Exam Syllabus for Chapter 2: Structure of Atom

Key Topics

  1. Discovery of Subatomic Particles
    • Electron, Proton, and Neutron
    • Experiments by J.J. Thomson and Rutherford
  2. Atomic Models
    • Thomson’s Model – The "Plum Pudding" model
    • Rutherford’s Nuclear Model – Gold foil experiment
    • Bohr’s Model of the Hydrogen Atom
  3. Dual Nature of Matter and Radiation
    • Photoelectric Effect – Einstein’s explanation
    • de Broglie’s Hypothesis – Matter waves
  4. Heisenberg’s Uncertainty Principle
    • Limitation in determining position and momentum simultaneously
  5. Quantum Mechanical Model of Atom
    • Introduction to Schrödinger’s Wave Equation
    • Concept of orbitals and shapes of s, p, and d orbitals
  6. Quantum Numbers
    • Principal (n), Azimuthal (l), Magnetic (m), and Spin (s)
    • Significance and rules of electron filling
  7. Electronic Configuration of Atoms
    • Aufbau Principle – Electrons fill lower energy orbitals first
    • Pauli’s Exclusion Principle – No two electrons can have identical quantum numbers
    • Hund’s Rule of Maximum Multiplicity
  8. Hydrogen Spectrum
    • Explanation of line spectra based on Bohr’s theory
  9. Limitations of Bohr’s Model
    • Transition to modern quantum mechanics

Tips to Master The Chapter Structure of Atom

  • Understand and visualize atomic models and orbitals for better retention.
  • Focus on the photoelectric effect and Heisenberg’s principle, as they are conceptually important.
  • Practice writing electronic configurations using rules like the Aufbau Principle and Hund’s Rule.
  • Solve questions on quantum numbers to strengthen your grasp of the concept.

💡 Structure of Atom forms the foundation for further chapters like Chemical Bonding and Periodic Properties. Mastering it will make future topics easier to understand.

Happy studying, and good luck with your preparations! 😊

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