Physics: Free Online MCQ Test on Atoms. Based Theory Questions With Answers

Atoms mcq class 12 with answers

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1. The first model of the atom in 1898 was proposed by:

A. Ernst Rutherford

B. Albert Einstein

C. J.J. Thomson

D. Niels Bohr

The correct answer is C. J.J. Thomson.

Explanation:
J.J. Thomson proposed the first model of the atom, known as the **"plum pudding model"**, in 1898. According to this model:

• The atom consists of a positively charged sphere with negatively charged electrons embedded within it, much like raisins in a pudding.
• This model explained the neutrality of atoms, as the positive and negative charges balanced each other.

However, this model could not explain the results of later experiments, such as the Geiger-Marsden scattering experiment conducted under Rutherford's guidance, which led to the discovery of the nucleus.

Key Points:

• J.J. Thomson's discovery of the electron in 1897 led to his atomic model.
• The model suggested that atoms are divisible, contrary to Dalton's atomic theory.

Conclusion:
The "plum pudding model" was an important milestone in atomic theory, providing the first conceptual framework for the structure of the atom, even though it was eventually replaced by more accurate models.

2. In the Geiger-Marsden scattering experiment, the trajectory traced by an α-particle depends on:

A. number of collisions

B. number of scattered α-particles

C. impact parameter

D. none of these

The correct answer is C. impact parameter.

Explanation:
The Geiger-Marsden scattering experiment, conducted under Rutherford's guidance, investigated the scattering of α-particles on a thin gold foil. The trajectory of the α-particles depended on the **impact parameter**, which is the perpendicular distance between the trajectory of the α-particle and the center of the nucleus.

• A smaller impact parameter results in a larger deflection angle because the α-particle passes closer to the nucleus and experiences a stronger electrostatic repulsion.
• A larger impact parameter results in minimal or no deflection as the α-particle is farther away from the nucleus.

Key Formula: The deflection angle \( \theta \) is inversely related to the impact parameter \( b \): \[ \theta \propto \frac{1}{b} \]
Conclusion:
The impact parameter determines how close an α-particle comes to the nucleus and, consequently, its trajectory. This insight was crucial for Rutherford's discovery of the nuclear structure of the atom.

3. In the Geiger-Marsden scattering experiment, the number of scattered particles detected is maximum and minimum at the scattering angles, respectively, at:

A. 0° and 180°

B. 180° and 0°

C. 90° and 180°

D. 45° and 90°

The correct answer is A. 0° and 180°.

Explanation:
In the Geiger-Marsden scattering experiment:

• Most of the α-particles pass straight through the gold foil without any deflection, corresponding to a scattering angle of \( 0^\circ \). This indicates that atoms are mostly empty space.
• Some α-particles are deflected at larger angles due to their close encounter with the positively charged nucleus, which causes strong electrostatic repulsion.
• The number of particles scattered at an angle of \( 180^\circ \) (directly backward) is minimal, as this occurs only in rare cases of direct head-on collision with the nucleus.

Key Insights:

• The distribution of scattering angles demonstrated the presence of a small, dense, positively charged nucleus at the center of the atom.
• Angles near \( 0^\circ \) have the maximum scattering particles, whereas \( 180^\circ \) represents a rare but significant observation.

Conclusion:
The observations from the experiment supported Rutherford's nuclear model of the atom, proving that most of the mass and positive charge of an atom is concentrated in a tiny nucleus.

4. In the Geiger-Marsden scattering experiment, in the case of a head-on collision, the impact parameter should be:

A. maximum

B. minimum

C. infinite

D. zero

The correct answer is D. zero.

Explanation:
The impact parameter is the perpendicular distance between the trajectory of an α-particle and the center of the nucleus. In a **head-on collision**:

• The α-particle moves directly toward the center of the nucleus without any lateral offset.
• This results in an impact parameter \( b = 0 \).

During a head-on collision:

• The α-particle experiences maximum electrostatic repulsion.
• This causes the α-particle to scatter backward, corresponding to a scattering angle \( \theta = 180^\circ \).

Key Insights:

• A zero impact parameter ensures direct interaction with the nucleus, which is a rare occurrence given the small size of the nucleus compared to the atom.
• This phenomenon supports the nuclear model, showing that the nucleus is compact and highly dense.

Conclusion:
For a head-on collision in the Geiger-Marsden experiment, the impact parameter must be zero, resulting in the α-particle being deflected at the maximum possible angle of \( 180^\circ \).

5. Rutherford’s experiments suggested that the size of the nucleus is about:

A. \(10^{-14} \, \text{m}\) to \(10^{-12} \, \text{m}\)

B. \(10^{-15} \, \text{m}\) to \(10^{-13} \, \text{m}\)

C. \(10^{-15} \, \text{m}\) to \(10^{-14} \, \text{m}\)

D. \(10^{-15} \, \text{m}\) to \(10^{-12} \, \text{m}\)

The correct answer is B. \(10^{-15} \, \text{m}\) to \(10^{-13} \, \text{m}\).

Explanation:
Rutherford’s scattering experiment revealed critical details about the size of the nucleus:

• The majority of α-particles passed through the gold foil without any deflection, indicating that atoms are mostly empty space.
• Only a few α-particles were deflected at large angles, which suggested the presence of a small, dense, positively charged core, i.e., the nucleus.

By analyzing the deflection of α-particles, Rutherford calculated that the diameter of the nucleus is approximately in the range of \(10^{-15} \, \text{m}\) to \(10^{-13} \, \text{m}\). This size is incredibly small compared to the overall size of the atom, which is approximately \(10^{-10} \, \text{m}\).

Key Points:

• The nucleus is tiny and contains nearly all the mass of the atom.
• The small size explains why most α-particles pass through without deflection.

Conclusion:
Rutherford’s experiments showed that the nucleus is extremely small, with a size range of \(10^{-15} \, \text{m}\) to \(10^{-13} \, \text{m}\), highlighting its compactness and density.

6. Which of the following spectral series falls within the visible range of electromagnetic radiation?

A. Lyman series

B. Balmer series

C. Paschen series

D. Pfund series

The correct answer is B. Balmer series.

Explanation:
The spectral series of the hydrogen atom correspond to the transitions of electrons between different energy levels. These series are categorized based on the final energy level to which the electron transitions:

• Lyman series: Transitions to \( n = 1 \); falls in the ultraviolet region.
• Balmer series: Transitions to \( n = 2 \); falls in the visible spectrum, producing visible light.
• Paschen series: Transitions to \( n = 3 \); falls in the infrared region.
• Pfund series: Transitions to \( n = 5 \); falls in the far-infrared region.

The Balmer series includes lines in the visible range (wavelengths approximately between 400 nm and 700 nm) and is named after Johann Balmer, who first derived the formula for these transitions.

Key Insight: The Balmer series is unique because it is the only spectral series of hydrogen whose wavelengths fall within the visible light spectrum.

Conclusion:
Among the options provided, the Balmer series is the only one that corresponds to the visible range of electromagnetic radiation.

7. The first spectral series was discovered by:

A. Balmer

B. Lyman

C. Paschen

D. Pfund

The correct answer is B. Lyman.

Explanation:
The spectral series of hydrogen were discovered in the following order:

• Lyman series: The first spectral series discovered in 1906 by Theodore Lyman. It corresponds to electron transitions to the \( n = 1 \) energy level and falls in the ultraviolet region.
• Balmer series: Discovered in 1885 by Johann Balmer, corresponding to transitions to \( n = 2 \), falling in the visible spectrum.
• Paschen series: Discovered by Friedrich Paschen in 1908, corresponding to transitions to \( n = 3 \), in the infrared region.
• Pfund series: Discovered by August Pfund in 1924, corresponding to transitions to \( n = 5 \), in the far-infrared region.

The Lyman series was the first spectral series discovered because it lies in the ultraviolet range, where early spectroscopic tools were first capable of detection. This discovery significantly advanced the understanding of atomic spectra and electron transitions.

Conclusion:
The Lyman series, discovered by Theodore Lyman, is the first spectral series identified in the hydrogen atom's emission spectrum.

8. Which of the following postulates of the Bohr model led to the quantization of energy of the hydrogen atom?

A. The electron goes around the nucleus in circular orbits.

B. The angular momentum of the electron can only be an integral multiple of \( \frac{h}{2\pi} \).

C. The magnitude of the linear momentum of the electron is quantized.

D. Quantization of energy is itself a postulate of the Bohr model.

The correct answer is B. The angular momentum of the electron can only be an integral multiple of \( \frac{h}{2\pi} \).

Explanation:
Bohr's model of the hydrogen atom introduced several revolutionary concepts, one of which was the quantization of energy levels. This was based on the following postulate:

• The angular momentum of an electron in a circular orbit around the nucleus is quantized and given by: \[ L = m_e v r = n \frac{h}{2\pi} \] where:
  • \( L \) is the angular momentum,
  • \( m_e \) is the mass of the electron,
  • \( v \) is the orbital velocity,
  • \( r \) is the radius of the orbit,
  • \( h \) is Planck's constant,
  • \( n \) is a positive integer (quantum number).

This quantization of angular momentum restricts the electron to specific orbits, each corresponding to a discrete energy level. The energy of the electron in the \( n \)-th orbit is: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] This explains why the energy of the hydrogen atom is quantized and why electrons can only occupy specific energy levels.

Conclusion:
The quantization of angular momentum in Bohr's model directly leads to the quantization of energy levels in the hydrogen atom, explaining the discrete spectral lines observed experimentally.

9. The Bohr model of atoms assumes that:

A. The angular momentum of electrons is quantized.

B. It uses Einstein’s photoelectric equation.

C. It predicts continuous emission spectra for atoms.

D. It predicts the same emission spectra for all types of atoms.

The correct answer is A. The angular momentum of electrons is quantized.

Explanation:
The Bohr model introduced several key postulates that differed from earlier atomic models:

• The model assumes that electrons move in specific orbits around the nucleus with quantized angular momentum: \[ L = m_e v r = n \frac{h}{2\pi} \] where \( n \) is an integer. This quantization prevents the continuous spectrum of wavelengths from being emitted by atoms, which is a characteristic of classical physics.
• The Bohr model does not use Einstein’s photoelectric equation, which is applicable to the emission of electrons from the surface of metals under the influence of light.
• It does not predict continuous emission spectra but rather discrete spectral lines corresponding to electron transitions between energy levels.
• The model specifically applies to hydrogen-like atoms and does not account for the complexities of multi-electron atoms.

Conclusion:
The key feature of the Bohr model is the quantization of angular momentum, which is why it successfully predicted the discrete spectral lines observed in hydrogen.

10. If \( n \) is the orbit number of the electron in a hydrogen atom, the correct statement among the following is:

A. Electron energy increases as \( n \) increases.

B. Hydrogen emits infrared rays for the electron transition from \( n = 3 \) to \( n = 1 \).

C. Electron energy is zero for \( n = 1 \)

D. Electron energy varies as \( n^2 \)

The correct answer is A. Electron energy increases as \( n \) increases.

Explanation:
In the Bohr model of the hydrogen atom, the energy levels of electrons are quantized and are represented by: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] where:

• \( E_n \) is the energy of the electron in the \( n \)-th orbit.
• \( n \) is the principal quantum number (orbit number).
• The negative sign indicates that the energy is lower than that of free electrons.

As \( n \) increases, the value of \( E_n \) becomes less negative, which means that the electron is in a higher energy level. Thus, the energy of the electron increases with increasing \( n \). This is why higher orbits correspond to greater energy levels.

Key Insights:

• Transitions between energy levels cause the emission or absorption of photons with specific energies.
• The largest energy difference is between the \( n = 3 \) and \( n = 1 \) orbits, which corresponds to ultraviolet light.

Conclusion:
The energy of an electron in a hydrogen atom is lowest when \( n = 1 \) and increases as \( n \) increases.

11. If the radius of the innermost electronic orbit of a hydrogen atom is \(5.3 \times 10^{-11} \, \text{m}\), then the radii of the \(n = 2\) orbit is:

A. 2.12 Å

B. 1.12 Å

C. 3.22 Å

D. 4.54 Å

The correct answer is A. 2.12 Å.

Explanation:
The radius of an orbit in the Bohr model of the hydrogen atom is given by: \[ r_n = n^2 \cdot r_1 \] where:

• \( r_n \) is the radius of the orbit in question,
• \( n \) is the principal quantum number (orbit number),
• \( r_1 \) is the radius of the first orbit (innermost orbit).

Given that \( r_1 = 5.3 \times 10^{-11} \, \text{m} \), for \( n = 2 \): \[ r_2 = 2^2 \cdot r_1 = 4 \cdot r_1 \] \[ r_2 = 4 \cdot (5.3 \times 10^{-11} \, \text{m}) \] \[ r_2 = 2.12 \times 10^{-10} \, \text{m} = 2.12 \, \text{Å} \] where 1 Å (angstrom) is \( 10^{-10} \, \text{m} \).

Conclusion:
The radius of the \( n = 2 \) orbit for a hydrogen atom is 2.12 Å, which represents the distance between the electron and the nucleus in this particular energy level.

12. The quantum number associated with the angular momentum of an electron in a hydrogen atom is:

A. \( l \)

B. \( n \)

C. \( m \)

D. \( s \)

The correct answer is A. \( l \).

**Explanation:** The quantum number \( l \) is associated with the angular momentum of an electron in an atom. It determines the shape and angular momentum of the electron’s orbital. The principal quantum number \( n \) determines the energy level and size of the orbital, while the magnetic quantum number \( m \) specifies the orientation of the orbital in space, and \( s \) denotes the spin quantum number. The angular momentum quantum number \( l \) can take values from 0 to \( n-1 \) for a given principal quantum number \( n \). Conclusion:
The angular momentum of an electron in a hydrogen atom is represented by the quantum number \( l \).

13. The spectral series corresponding to the transition of an electron from \( n = 3 \) to \( n = 1 \) in a hydrogen atom is:

A. Lyman series

B. Balmer series

C. Paschen series

D. Pfund series

The correct answer is A. Lyman series.

**Explanation:** The Lyman series corresponds to transitions to the \( n = 1 \) energy level in a hydrogen atom. The photon emitted in these transitions lies in the ultraviolet region of the electromagnetic spectrum. When an electron falls from \( n = 3 \) to \( n = 1 \), it releases energy corresponding to ultraviolet light. This makes it a transition within the Lyman series. Conclusion:
The transition from \( n = 3 \) to \( n = 1 \) in hydrogen results in emission within the Lyman series.

14. Which of the following transitions in a hydrogen atom results in the absorption of a photon?

A. \( n = 2 \) to \( n = 3 \)

B. \( n = 3 \) to \( n = 2 \)

C. \( n = 1 \) to \( n = 2 \)

D. \( n = 1 \) to \( n = 3 \)

The correct answer is A. \( n = 2 \) to \( n = 3 \).

**Explanation:** In the Bohr model of the hydrogen atom, an electron absorbs energy when it moves from a lower energy level to a higher one (e.g., from \( n = 2 \) to \( n = 3 \)). This process corresponds to the absorption of a photon with energy equal to the difference between these energy levels. Conversely, the emission of a photon occurs when an electron falls from a higher to a lower energy level. Conclusion:
The transition \( n = 2 \) to \( n = 3 \) corresponds to an absorption process, where the electron requires energy to move to a higher orbit.

15. According to the Heisenberg Uncertainty Principle, which pair of physical properties cannot be simultaneously measured with arbitrary precision?

A. Position and momentum

B. Energy and time

C. Angular momentum and energy

D. Charge and mass

The correct answer is A. Position and momentum.

**Explanation:** The Heisenberg Uncertainty Principle states that it is impossible to simultaneously know both the exact position (\( x \)) and momentum (\( p \)) of a particle with absolute precision. The more precisely one of these quantities is known, the less precisely the other can be known. This is expressed mathematically by the inequality: \[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \] where \( \Delta x \) is the uncertainty in position, \( \Delta p \) is the uncertainty in momentum, and \( h \) is Planck's constant. Conclusion:
The Heisenberg Uncertainty Principle fundamentally limits the precision with which certain pairs of physical properties, such as position and momentum, can be measured.