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Structure of Atom

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Summary

Summary of Structure of Atom

  • Atoms are the smallest parts of an element that chemically react.
  • John Dalton proposed the first atomic theory in 1808, viewing atoms as indivisible particles.
  • Experimental evidence in the late 19th century showed that atoms are divisible and consist of:
    • Electrons
    • Protons
    • Neutrons
  • Thomson's model (1898): Atom as a uniform sphere of positive charge with electrons embedded.
  • Rutherford's model (1909): Atom has a tiny positively charged nucleus with electrons orbiting around it.
  • Bohr's model (1913): Electrons move in fixed circular orbits around the nucleus, with quantized energy levels.
  • Schrödinger's equation (1926): Describes electron distributions and energy levels, incorporating wave-particle duality.
  • Quantum mechanical model: Electrons are distributed in shells, subshells, and orbitals, with energy levels defined by quantum numbers.
  • Pauli exclusion principle: No two electrons can have the same set of four quantum numbers.
  • Hund's rule: Electrons fill orbitals singly before pairing up.

Learning Objectives

Learning Objectives

  • Know about the discovery of electron, proton, and neutron and their characteristics.
  • Describe Thomson, Rutherford, and Bohr atomic models.
  • Understand the important features of the quantum mechanical model of the atom.
  • Understand the nature of electromagnetic radiation and Planck's quantum theory.
  • Explain the photoelectric effect and describe features of atomic spectra.
  • State the de Broglie relation and Heisenberg uncertainty principle.
  • Define an atomic orbital in terms of quantum numbers.
  • State the Aufbau principle, Pauli exclusion principle, and Hund's rule of maximum multiplicity.
  • Write the electronic configurations of atoms.

Detailed Notes

Structure of Atom

Objectives

  • Understand the discovery of electron, proton, and neutron and their characteristics.
  • Describe Thomson, Rutherford, and Bohr atomic models.
  • Understand the important features of the quantum mechanical model of the atom.
  • Understand the nature of electromagnetic radiation and Planck's quantum theory.
  • Explain the photoelectric effect and describe features of atomic spectra.
  • State the de Broglie relation and Heisenberg uncertainty principle.
  • Define an atomic orbital in terms of quantum numbers.
  • State the Aufbau principle, Pauli exclusion principle, and Hund's rule of maximum multiplicity.
  • Write the electronic configurations of atoms.

Historical Background

  • The concept of atoms dates back to early Indian and Greek philosophers (400 B.C.), who believed atoms were the fundamental building blocks of matter.
  • John Dalton proposed the atomic theory in 1808, which regarded atoms as indivisible particles.

Atomic Models

Thomson Model (1898)

  • Proposed that an atom consists of a uniform sphere of positive electricity with electrons embedded in it.

Rutherford Model (1909)

  • Concluded that atoms have a tiny positively charged nucleus with electrons revolving around it in circular orbits.
  • This model improved upon Thomson's but could not explain atomic stability.

Bohr Model (1913)

  • Proposed that electrons move in circular orbits around the nucleus, with specific energy levels.
  • Could explain hydrogen atom spectra but not multi-electron atoms.

Quantum Mechanical Model

  • Erwin Schrödinger (1926) proposed an equation to describe electron distributions and energy levels.
  • Incorporates wave-particle duality and is consistent with the Heisenberg uncertainty principle.
  • Electrons are distributed in shells, subshells, and orbitals.

Key Principles

  • Aufbau Principle: Electrons fill orbitals starting from the lowest energy level.
  • Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers.
  • Hund's Rule: Electrons will singly occupy degenerate orbitals before pairing up.

Electronic Configurations

ElementConfiguration
H1s¹
He1s²
Li1s² 2s¹
Be1s² 2s²
B1s² 2s² 2p¹
C1s² 2s² 2p²
N1s² 2s² 2p³
O1s² 2s² 2p⁴
F1s² 2s² 2p⁵
Ne1s² 2s² 2p⁶
......

Conclusion

  • The structure of the atom is fundamental to understanding chemical behavior and reactions.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

  • Misunderstanding Atomic Models: Students often confuse the characteristics of Thomson, Rutherford, and Bohr models. Ensure you can describe each model's key features and limitations.
  • Ignoring Quantum Principles: Failing to apply the Heisenberg uncertainty principle and the concept of wave-particle duality can lead to incorrect conclusions about electron behavior.
  • Incorrect Electron Configuration: Be cautious with writing electronic configurations, especially for transition metals and elements with exceptional configurations like Cr and Cu.
  • Neglecting Effective Nuclear Charge: Students may overlook how effective nuclear charge affects electron behavior in different orbitals. Understand which electrons experience higher or lower effective nuclear charge.

Tips for Success

  • Review Key Definitions: Make sure you understand definitions such as atomic orbital, quantum numbers, and the principles of electron configuration (Aufbau, Pauli exclusion, Hund's rule).
  • Practice Calculations: Work on problems involving calculations of energy, wavelength, and frequency to solidify your understanding of electromagnetic radiation.
  • Understand the Historical Context: Knowing the historical development of atomic theory can help you appreciate the significance of various models and their limitations.
  • Use Visual Aids: Diagrams of atomic models and electron configurations can help visualize concepts better. Practice sketching these out.
  • Memorize Key Formulas: Familiarize yourself with formulas related to energy, wavelength, and frequency, as these are frequently tested in exams.

Practice & Assessment

Multiple Choice Questions

A.

n = 5 to n = 4

B.

n = 4 to n = 3

C.

n = 3 to n = 2

D.

n = 2 to n = 1
Correct Answer: D

Solution:

The energy of a photon emitted during a transition is given by the difference in energy levels. The transition from n = 2 to n = 1 releases the most energy as it involves the largest difference in energy levels.

A.

2

B.

3

C.

4

D.

5
Correct Answer: C

Solution:

For a given principal quantum number nn, the number of subshells is equal to nn. Therefore, for n=4n = 4, there are 4 subshells.

A.

Electron in 2p orbital

B.

Electron in 3p orbital

C.

Electron in 4p orbital

D.

Electron in 1s orbital
Correct Answer: C

Solution:

The electron in the 4p orbital experiences the lowest effective nuclear charge due to increased shielding by inner electrons.

A.

1.56 \times 10^3

B.

1.56 \times 10^2

C.

1.56 \times 10^4

D.

1.56 \times 10^5
Correct Answer: A

Solution:

The charge of one electron is approximately 1.6 \times 10^{-19} C. Therefore, the number of electrons is 2.5 \times 10^{-16} C / 1.6 \times 10^{-19} C/electron = 1.56 \times 10^3.

A.

Principal quantum number (nn)

B.

Azimuthal quantum number (ll)

C.

Magnetic quantum number (mlm_l)

D.

Spin quantum number (msm_s)
Correct Answer: C

Solution:

The magnetic quantum number (mlm_l) gives information about the spatial orientation of the orbital.

A.

2

B.

6

C.

10

D.

14
Correct Answer: C

Solution:

The 3d subshell can hold a maximum of 10 electrons. This is because there are 5 orbitals in the d subshell, and each orbital can hold 2 electrons.

A.

Positive

B.

Negative

C.

Neutral

D.

Depends on the isotope
Correct Answer: C

Solution:

Neutrons are neutral particles, meaning they have no charge.

A.

n = 3 to n = 2

B.

n = 4 to n = 3

C.

n = 2 to n = 1

D.

n = 5 to n = 4
Correct Answer: C

Solution:

The transition from n=2n = 2 to n=1n = 1 in the hydrogen atom corresponds to the Lyman series, which is in the ultraviolet region of the electromagnetic spectrum.

A.

Dalton's atomic model

B.

Thomson's atomic model

C.

Rutherford's atomic model

D.

Bohr's atomic model
Correct Answer: A

Solution:

Dalton's atomic model regarded the atom as the ultimate particle of matter.

A.

n=3n = 3 to n=1n = 1

B.

n=5n = 5 to n=2n = 2

C.

n=6n = 6 to n=3n = 3

D.

n=4n = 4 to n=3n = 3
Correct Answer: A

Solution:

The transition n=3n = 3 to n=1n = 1 in the hydrogen spectrum has the same wavelength as the Balmer transition n=4n = 4 to n=2n = 2 of He+^+ spectrum.

A.

0

B.

1

C.

2

D.

3
Correct Answer: C

Solution:

The principal quantum number nn for the second shell is 2.

A.

0.33 nm

B.

0.15 nm

C.

0.25 nm

D.

0.18 nm
Correct Answer: A

Solution:

The de Broglie wavelength is given by λ=hmv\lambda = \frac{h}{mv}, where hh is Planck's constant (6.626×10346.626 \times 10^{-34} Js), mm is the mass of the electron (9.109×10319.109 \times 10^{-31} kg), and vv is the velocity. Substituting these values, λ=6.626×10349.109×1031×2.19×106=0.33 nm\lambda = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 2.19 \times 10^6} = 0.33 \text{ nm}.

A.

Principal quantum number (nn)

B.

Azimuthal quantum number (ll)

C.

Magnetic quantum number (mlm_l)

D.

Spin quantum number (msm_s)
Correct Answer: B

Solution:

The azimuthal quantum number (ll) is used to describe the shape of an orbital.

A.

Principal quantum number (n)

B.

Azimuthal quantum number (l)

C.

Magnetic quantum number (m_l)

D.

Spin quantum number (m_s)
Correct Answer: B

Solution:

The azimuthal quantum number (l) determines the shape of the atomic orbital, distinguishing between s, p, d, and f orbitals.

A.

It states that the position and velocity of a particle can be measured exactly at the same time.

B.

It states that the position and momentum of a particle cannot both be precisely determined at the same time.

C.

It states that the energy and frequency of a photon cannot be simultaneously known.

D.

It states that the charge and mass of an electron cannot be simultaneously known.
Correct Answer: B

Solution:

The Heisenberg Uncertainty Principle asserts that it is impossible to simultaneously determine with arbitrary precision both the position and momentum of a particle.

A.

Determining the shape of the orbital

B.

Determining the size and energy of the orbital

C.

Determining the orientation of the orbital

D.

Determining the charge of the electron
Correct Answer: B

Solution:

The principal quantum number 'n' determines the size and to a large extent the energy of the orbital.

A.

Shape of the orbital

B.

Size and energy of the orbital

C.

Orientation of the orbital

D.

Spin of the electron
Correct Answer: B

Solution:

The principal quantum number nn determines the size and to a large extent the energy of the orbital.

A.

Dalton's atomic theory

B.

Thomson's atomic model

C.

Rutherford's atomic model

D.

Bohr's atomic model
Correct Answer: A

Solution:

John Dalton proposed the atomic theory which regarded the atom as the ultimate particle of matter.

A.

n = 4 to n = 3

B.

n = 3 to n = 2

C.

n = 2 to n = 1

D.

n = 5 to n = 4
Correct Answer: C

Solution:

The energy difference between levels decreases as nn increases. The transition from n=2n = 2 to n=1n = 1 releases the most energy as it is the largest energy gap.

A.

n=5 to n=3

B.

n=6 to n=4

C.

n=3 to n=1

D.

n=2 to n=1
Correct Answer: A

Solution:

The wavelength of a transition is given by the Rydberg formula. For He+, the transition n=4 to n=2 corresponds to the same energy difference as n=5 to n=3 in hydrogen due to the difference in nuclear charge.

A.

0

B.

1

C.

2

D.

3
Correct Answer: B

Solution:

The azimuthal quantum number ll for a p orbital is 1.

A.

Plum pudding model

B.

Nuclear model

C.

Quantum mechanical model

D.

Bohr's model
Correct Answer: B

Solution:

Rutherford proposed the nuclear model of the atom, where a small, dense nucleus is surrounded by orbiting electrons.

A.

Intensity of the incident light

B.

Frequency of the incident light

C.

Work function of the metal

D.

Wavelength of the incident light
Correct Answer: A

Solution:

The kinetic energy of the emitted photoelectrons in the photoelectric effect is determined by the frequency of the incident light and the work function of the metal. The intensity of the incident light affects the number of photoelectrons emitted, not their kinetic energy.

A.

n = 3, l = 2, m_l = 3, m_s = +1/2

B.

n = 2, l = 1, m_l = 0, m_s = -1/2

C.

n = 4, l = 3, m_l = -3, m_s = +1/2

D.

n = 1, l = 0, m_l = 0, m_s = -1/2
Correct Answer: A

Solution:

The magnetic quantum number mlm_l can have values from l-l to +l+l. For l=2l = 2, mlm_l can only be -2, -1, 0, 1, or 2. Thus, ml=3m_l = 3 is not possible.

A.

Protons: 20, Neutrons: 20, Electrons: 18

B.

Protons: 20, Neutrons: 20, Electrons: 20

C.

Protons: 18, Neutrons: 22, Electrons: 18

D.

Protons: 18, Neutrons: 22, Electrons: 20
Correct Answer: A

Solution:

The atomic number Z=20Z = 20 indicates that there are 20 protons in the atom. The mass number A=40A = 40 means there are 40 - 20 = 20 neutrons. Losing two electrons results in 20 - 2 = 18 electrons in the cation.

A.

The nucleus is extremely small and dense.

B.

The atoms of the element have no nucleus.

C.

The nucleus is positively charged.

D.

The nucleus is very large and occupies most of the atomic volume.
Correct Answer: B

Solution:

If no deflection is observed, it suggests that there is no dense, positively charged nucleus to cause the deflection of alpha particles, implying the atoms might lack a nucleus.

A.

It could not explain the stability of atoms.

B.

It failed to account for the existence of isotopes.

C.

It did not include the concept of quantized energy levels.

D.

It was unable to explain the photoelectric effect.
Correct Answer: A

Solution:

The Rutherford model could not explain the stability of atoms as it suggested that electrons moving in orbits would radiate energy and spiral into the nucleus.

A.

s

B.

p

C.

d

D.

f
Correct Answer: C

Solution:

The azimuthal quantum number l=2l = 2 corresponds to the 'd' sub-shell.

A.

The kinetic energy of ejected electrons is independent of the frequency of incident light.

B.

The number of ejected electrons is proportional to the intensity of incident light.

C.

The threshold frequency is dependent on the intensity of incident light.

D.

The photoelectric effect can occur with any frequency of light if the intensity is high enough.
Correct Answer: B

Solution:

The number of ejected electrons is proportional to the intensity of the incident light, while the kinetic energy of the electrons depends on the frequency of the light.

A.

Sulfur

B.

Chlorine

C.

Phosphorus

D.

Argon
Correct Answer: A

Solution:

Sulfur has an atomic number of 16.

A.

It can be emitted in continuous waves.

B.

It can be emitted in discrete packets of energy called quanta.

C.

It cannot be emitted at all.

D.

It is always visible to the human eye.
Correct Answer: B

Solution:

Planck's quantum theory states that electromagnetic radiation is emitted in discrete packets of energy called quanta.

A.

1.33 x 10^9

B.

1.33 x 10^10

C.

1.33 x 10^11

D.

1.33 x 10^12
Correct Answer: B

Solution:

To find the number of carbon atoms, divide 20 cm (200,000,000 nm) by the diameter of a carbon atom (0.15 nm): 200,000,000 nm / 0.15 nm = 1.33 x 10^10.

A.

Ultraviolet

B.

Visible

C.

Infrared

D.

Microwave
Correct Answer: B

Solution:

The transition from n=5 to n=2 in the hydrogen atom corresponds to the Balmer series, which emits visible light.

A.

3p electrons are closer to the nucleus than 3s electrons.

B.

3p electrons experience more shielding than 3s electrons.

C.

3p electrons experience less shielding than 3s electrons.

D.

3p electrons have a higher principal quantum number than 3s electrons.
Correct Answer: C

Solution:

The effective nuclear charge is higher for 3p electrons because they experience less shielding from inner electrons compared to 3s electrons, which are closer to the nucleus and shielded more effectively.

A.

Fe^3+

B.

Co^3+

C.

Mn^3+

D.

Ni^3+
Correct Answer: A

Solution:

Let the number of electrons be xx. Then, the number of neutrons is x+0.304x=1.304xx + 0.304x = 1.304x. The number of protons is x+3x + 3. Thus, x+1.304x+x+3=56x + 1.304x + x + 3 = 56. Solving gives x=23x = 23. Therefore, the number of protons is 2626, which corresponds to iron (Fe). The ion is Fe3+\text{Fe}^{3+}.

A.

Divisible

B.

Indivisible

C.

Made of electrons

D.

Made of protons
Correct Answer: B

Solution:

Dalton's atomic theory proposed that atoms are indivisible particles, which are the fundamental building blocks of matter.

A.

Dalton's atomic model

B.

Thomson's atomic model

C.

Rutherford's atomic model

D.

Bohr's atomic model
Correct Answer: C

Solution:

Rutherford's atomic model is likened to a small scale solar system where the nucleus is the sun and electrons are the planets.

A.

n = 3 to n = 2

B.

n = 4 to n = 3

C.

n = 2 to n = 1

D.

n = 5 to n = 4
Correct Answer: C

Solution:

The Lyman series corresponds to transitions where the final state is n=1n = 1. Thus, the transition from n=2n = 2 to n=1n = 1 is part of the Lyman series.

A.

8

B.

10

C.

12

D.

14
Correct Answer: A

Solution:

The charge of one electron is approximately -1.602 x 10^-19 C. The number of electrons is calculated by dividing the total charge by the charge of one electron: -1.282 x 10^-18 C / -1.602 x 10^-19 C/electron ≈ 8 electrons.

A.

It explained the electrical nature of matter.

B.

It stated that atoms are divisible.

C.

It explained the law of conservation of mass.

D.

It proposed the existence of sub-atomic particles.
Correct Answer: C

Solution:

Dalton's atomic theory successfully explained the law of conservation of mass.

A.

8×1038 \times 10^3

B.

1.5×1041.5 \times 10^4

C.

3×1043 \times 10^4

D.

5×1045 \times 10^4
Correct Answer: A

Solution:

Energy of one photon E=hcλE = \frac{hc}{\lambda}, where h=6.626×1034 Jsh = 6.626 \times 10^{-34} \text{ Js}, c=3×108 m/sc = 3 \times 10^8 \text{ m/s}, and λ=600×109 m\lambda = 600 \times 10^{-9} \text{ m}. E=3.31×1019 JE = 3.31 \times 10^{-19} \text{ J}. Number of photons = 3.15×10183.31×10198×103\frac{3.15 \times 10^{-18}}{3.31 \times 10^{-19}} \approx 8 \times 10^3.

A.

Thomson's model

B.

Rutherford's model

C.

Bohr's model

D.

Dalton's model
Correct Answer: B

Solution:

Rutherford's model describes the atom as having a central nucleus with electrons orbiting around it, similar to planets orbiting the sun.

A.

Plum pudding model

B.

Nuclear model

C.

Planetary model

D.

Quantum mechanical model
Correct Answer: A

Solution:

J.J. Thomson proposed the plum pudding model, which depicted the atom as a sphere of positive charge with electrons embedded in it.

A.

1s² 2s² 2p⁶ 3s² 3p³

B.

1s² 2s² 2p⁶ 3s² 3p⁶

C.

1s² 2s² 2p⁶ 3s² 3p⁴

D.

1s² 2s² 2p⁶ 3s² 3p⁵
Correct Answer: B

Solution:

The element with atomic number 15 is phosphorus (P), which has the electronic configuration 1s² 2s² 2p⁶ 3s² 3p³. When it gains three electrons, it becomes P³⁻ with the configuration 1s² 2s² 2p⁶ 3s² 3p⁶.

A.

Energy is emitted continuously.

B.

Energy is emitted in discrete packets called quanta.

C.

Energy is inversely proportional to frequency.

D.

Energy is directly proportional to wavelength.
Correct Answer: B

Solution:

Planck's quantum theory states that energy is emitted in discrete packets called quanta.

A.

3.32 W

B.

2.76 W

C.

4.12 W

D.

5.24 W
Correct Answer: A

Solution:

The energy of one photon is given by E=hcλE = \frac{hc}{\lambda}, where h=6.626×1034 J sh = 6.626 \times 10^{-34} \text{ J s} and c=3×108 m/sc = 3 \times 10^8 \text{ m/s}. Substituting λ=337.1×109 m\lambda = 337.1 \times 10^{-9} \text{ m}, we find E=5.89×1019 JE = 5.89 \times 10^{-19} \text{ J}. The total energy is 5.89×1019 J×5.6×1024=3.30×106 J5.89 \times 10^{-19} \text{ J} \times 5.6 \times 10^{24} = 3.30 \times 10^6 \text{ J}. Assuming the emission is continuous over 1 second, the power is 3.30×106 W3.30 \times 10^6 \text{ W}.

A.

It is possible to know both the position and momentum of a particle with absolute certainty.

B.

The position and momentum of a particle can be measured simultaneously with arbitrary precision.

C.

The more precisely the position of a particle is known, the less precisely its momentum can be known.

D.

The uncertainty principle applies only to macroscopic objects.
Correct Answer: C

Solution:

The Heisenberg Uncertainty Principle states that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa. This principle is a fundamental concept in quantum mechanics.

A.

1.33 x 10^9

B.

1.33 x 10^10

C.

1.33 x 10^11

D.

1.33 x 10^12
Correct Answer: B

Solution:

The number of carbon atoms is calculated by dividing the total length by the diameter of one atom: 20 cm0.15 nm=1.33×1010\frac{20 \text{ cm}}{0.15 \text{ nm}} = 1.33 \times 10^{10}.

A.

Sulfur (S)

B.

Iron (Fe)

C.

Chromium (Cr)

D.

Krypton (Kr)
Correct Answer: C

Solution:

Chromium (Cr) has the electronic configuration [Ar] 3d5^5 4s1^1, with 6 unpaired electrons, which is the highest among the given options.

A.

Transition from n=4n=4 to n=2n=2 in the hydrogen atom

B.

Transition from n=6n=6 to n=4n=4 in the hydrogen atom

C.

Transition from n=3n=3 to n=1n=1 in the hydrogen atom

D.

Transition from n=5n=5 to n=2n=2 in the hydrogen atom
Correct Answer: A

Solution:

The energy levels of hydrogen and He+^+ are related by the Rydberg formula. The transition from n=5n=5 to n=3n=3 in He+^+ corresponds to the same energy as the transition from n=4n=4 to n=2n=2 in hydrogen due to the difference in nuclear charge.

A.

Dalton's atomic model

B.

Thomson's atomic model

C.

Rutherford's atomic model

D.

Bohr's atomic model
Correct Answer: C

Solution:

Rutherford's atomic model is often compared to a small scale solar system, with electrons orbiting a central nucleus.

A.

[Ne] 3s^2 3p^6

B.

[Ar]

C.

[Ne] 3s^2 3p^4

D.

[Ne] 3s^2 3p^6 4s^2
Correct Answer: B

Solution:

Element X with atomic number 20 has the electronic configuration [Ar] 4s^2. By losing two electrons, it forms a cation with the configuration [Ar].

A.

Br

B.

Kr

C.

Se

D.

Cl
Correct Answer: A

Solution:

Let the number of protons be pp. Neutrons are 31.7% more than protons, so neutrons = 1.317p1.317p. The mass number A=p+1.317p=81A = p + 1.317p = 81. Solving gives p35p \approx 35, which corresponds to bromine (Br).

A.

1

B.

2

C.

3

D.

4
Correct Answer: C

Solution:

The 3d subshell is part of the third shell, which corresponds to the principal quantum number n=3n = 3.

A.

2

B.

6

C.

10

D.

14
Correct Answer: B

Solution:

For n=4n=4, the possible values of ll are 0, 1, 2, and 3. For ml=1m_l = -1, electrons can be in the pp, dd, or ff subshells, which can each hold 2 electrons with ml=1m_l = -1. Therefore, the total is 6 electrons.

A.

Principal quantum number (nn)

B.

Azimuthal quantum number (ll)

C.

Magnetic quantum number (mlm_l)

D.

Spin quantum number (msm_s)
Correct Answer: C

Solution:

The magnetic quantum number (mlm_l) is associated with the orientation of an atomic orbital in space. It determines the number of orbitals and their orientation within a subshell.

A.

Br

B.

Kr

C.

I

D.

Xe
Correct Answer: B

Solution:

Let the number of protons be pp. Then the number of neutrons is 1.317p1.317p. The mass number is p+1.317p=81p + 1.317p = 81. Solving gives p=35p = 35, which corresponds to bromine (Br).

A.

More alpha particles will be deflected at large angles.

B.

Fewer alpha particles will be deflected at large angles.

C.

The deflection pattern will remain unchanged.

D.

Alpha particles will pass straight through without deflection.
Correct Answer: B

Solution:

Aluminum has a lower atomic number than gold, meaning its nucleus is smaller and has less positive charge. Thus, fewer alpha particles will be deflected at large angles compared to gold.

A.

Aufbau principle

B.

Pauli exclusion principle

C.

Hund's rule

D.

Heisenberg uncertainty principle
Correct Answer: B

Solution:

The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers.

A.

2.47 x 10^15 Hz

B.

3.28 x 10^15 Hz

C.

1.54 x 10^15 Hz

D.

4.57 x 10^15 Hz
Correct Answer: A

Solution:

The frequency can be calculated using the formula ν=RHc(1n121n22)\nu = R_H c \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right). Substituting the values, ν=1.097x107×3x108(112132)=2.47x1015 Hz\nu = 1.097 x 10^7 \times 3 x 10^8 \left(\frac{1}{1^2} - \frac{1}{3^2}\right) = 2.47 x 10^15 \text{ Hz}.

A.

The Bohr model describes electrons in fixed orbits, while the quantum mechanical model describes electrons in probabilistic orbitals.

B.

The Bohr model includes the concept of electron spin, while the quantum mechanical model does not.

C.

The Bohr model accounts for the Heisenberg uncertainty principle, while the quantum mechanical model does not.

D.

The Bohr model is based on quantum numbers, while the quantum mechanical model is not.
Correct Answer: A

Solution:

The Bohr model describes electrons moving in fixed circular orbits around the nucleus, similar to planets around the sun. In contrast, the quantum mechanical model describes electrons as existing in orbitals, which are regions of space where there is a high probability of finding an electron. This model incorporates the Heisenberg uncertainty principle, which states that one cannot simultaneously know the exact position and momentum of an electron.

A.

1

B.

2

C.

3

D.

4
Correct Answer: C

Solution:

For a given principal quantum number nn, the azimuthal quantum number ll can take values from 0 to n1n-1. Therefore, for n=3n=3, ll can be 0, 1, or 2, which gives three possible values.

A.

Dalton's atomic model

B.

Thomson's atomic model

C.

Rutherford's atomic model

D.

Bohr's atomic model
Correct Answer: C

Solution:

Rutherford's atomic model was the first to propose that atoms are made of sub-atomic particles, differing from Dalton's model.

A.

1

B.

2

C.

3

D.

4
Correct Answer: C

Solution:

For n = 3, there are three subshells: 3s, 3p, and 3d.

A.

No deflection of alpha particles

B.

Greater deflection of alpha particles

C.

Less deflection of alpha particles

D.

Complete absorption of alpha particles
Correct Answer: C

Solution:

Using a thin foil of light atoms like aluminum would result in less deflection of alpha particles compared to heavy atoms like gold.

A.

1s22s22p63s23p61s^2 2s^2 2p^6 3s^2 3p^6

B.

1s22s22p63s23p64s21s^2 2s^2 2p^6 3s^2 3p^6 4s^2

C.

1s22s22p63s23p63d21s^2 2s^2 2p^6 3s^2 3p^6 3d^2

D.

1s22s22p63s23p64s11s^2 2s^2 2p^6 3s^2 3p^6 4s^1
Correct Answer: A

Solution:

The element with atomic number 20 is calcium, with a neutral configuration of 1s22s22p63s23p64s21s^2 2s^2 2p^6 3s^2 3p^6 4s^2. Losing two electrons results in 1s22s22p63s23p61s^2 2s^2 2p^6 3s^2 3p^6.

A.

n = 3

B.

n = 4

C.

n = 5

D.

n = 6
Correct Answer: B

Solution:

The 4f subshell is part of the fourth shell, so the principal quantum number is n = 4.

A.

3

B.

9

C.

18

D.

27
Correct Answer: B

Solution:

For n=3n=3, the possible values of ll are 0, 1, and 2, corresponding to the ss, pp, and dd subshells respectively. The number of orbitals is given by n2=32=9n^2 = 3^2 = 9.

A.

8

B.

16

C.

32

D.

2
Correct Answer: B

Solution:

For n=4n = 4, there are 16 orbitals, and each orbital can hold 2 electrons with opposite spins. Therefore, 16 electrons can have ms=12m_s = -\frac{1}{2}.

A.

Principal quantum number (nn)

B.

Azimuthal quantum number (ll)

C.

Magnetic quantum number (mlm_l)

D.

Spin quantum number (msm_s)
Correct Answer: A

Solution:

The principal quantum number (nn) determines the size and energy of the orbital.

A.

1.1 eV

B.

2.1 eV

C.

3.1 eV

D.

4.1 eV
Correct Answer: C

Solution:

The energy of the photon is given by E = hc/λ. For λ = 400 nm, E = 3.1 eV. The work function (Φ) is the difference between the photon energy and the kinetic energy of the ejected electron: Φ = 3.1 eV - 2.0 eV = 1.1 eV.

A.

0.82×1018 J0.82 \times 10^{-18} \text{ J}

B.

1.64×1018 J1.64 \times 10^{-18} \text{ J}

C.

3.27×1018 J3.27 \times 10^{-18} \text{ J}

D.

6.54×1018 J6.54 \times 10^{-18} \text{ J}
Correct Answer: A

Solution:

The energy required to ionize an electron from the second orbit is given by E2=E1n2=3.27×101822=0.82×1018 JE_2 = \frac{E_1}{n^2} = \frac{3.27 \times 10^{-18}}{2^2} = 0.82 \times 10^{-18} \text{ J}.

A.

2

B.

6

C.

10

D.

14
Correct Answer: A

Solution:

For n=3n = 3 and l=0l = 0, there is one ss orbital which can hold 2 electrons.

A.

Electrons are indivisible particles.

B.

Electrons have a positive charge.

C.

Electrons are located in the nucleus.

D.

Electrons have the same mass as protons.
Correct Answer: A

Solution:

Early atomic models, such as Dalton's, considered electrons as fundamental, indivisible particles.

A.

Neutral

B.

1+

C.

2-

D.

2+
Correct Answer: C

Solution:

The ion has 2 more electrons than protons, giving it a charge of 2-.

A.

8

B.

18

C.

32

D.

10
Correct Answer: B

Solution:

The n=3 shell can have three subshells: 3s, 3p, and 3d. The maximum number of electrons is calculated as 2n^2 = 2(3)^2 = 18.

True or False

Correct Answer: True

Solution:

The principal quantum number 'n' determines the size and to a large extent the energy of the orbital.

Correct Answer: True

Solution:

Rutherford's model suggested that electrons orbit the nucleus, but according to electromagnetic theory, such motion would cause electrons to emit radiation and lose energy, leading to instability, which was not observed.

Correct Answer: False

Solution:

The principal quantum number 'n' determines the size and energy of the orbital, not its shape. The shape is determined by the azimuthal quantum number 'l'.

Correct Answer: True

Solution:

The nuclear model suggests that electrons move around the nucleus in well-defined orbits, similar to how planets orbit the sun.

Correct Answer: False

Solution:

Using light atoms like aluminum would result in different scattering patterns due to their lower atomic mass and charge.

Correct Answer: True

Solution:

In Milikan's experiment, X-rays were used to obtain the static electric charge on the oil drops.

Correct Answer: False

Solution:

The early Indian and Greek philosophers' ideas about atoms were mere speculations and there was no way to test them experimentally at that time.

Correct Answer: True

Solution:

The excerpt states that the rich diversity of chemical behavior of different elements can be traced to the differences in the internal structure of atoms of these elements.

Correct Answer: True

Solution:

Milikan's experiment involved obtaining the static electric charge on oil drops by shining X-rays.

Correct Answer: True

Solution:

The Bohr model postulates that electrons move in fixed orbits around the nucleus, similar to planets orbiting the sun, which was a significant advancement in atomic theory.

Correct Answer: False

Solution:

Dalton's atomic theory could not explain the results of experiments where substances like glass or ebonite get electrically charged when rubbed with silk or fur.

Correct Answer: True

Solution:

Rutherford's nuclear model of the atom is likened to a small-scale solar system, with the nucleus playing the role of the massive sun and electrons similar to the lighter planets.

Correct Answer: False

Solution:

Electrons in the 2p orbital experience a higher effective nuclear charge than those in the 4p orbital due to being closer to the nucleus and having fewer electron shielding effects.

Correct Answer: False

Solution:

The principal quantum number 'n' determines the size and energy of the orbital, not its shape. The azimuthal quantum number 'l' determines the shape.

Correct Answer: True

Solution:

The Rutherford model could not explain the stability of atoms because, according to classical mechanics and electromagnetic theory, electrons in orbit should emit radiation and spiral into the nucleus.

Correct Answer: True

Solution:

John Dalton, a British school teacher, proposed the atomic theory of matter in 1808, which regarded the atom as the ultimate particle of matter.

Correct Answer: False

Solution:

Rutherford's model could not explain the stability of the atom because it suggested that electrons, while moving in orbits, should emit radiation and spiral into the nucleus, which does not occur.

Correct Answer: False

Solution:

The word 'atom' is derived from the Greek word 'a-tomio', which means 'uncut-able' or 'non-divisible'.

Correct Answer: False

Solution:

The Rutherford model suggests that electrons move in well-defined orbits around the nucleus, not stationary.

Correct Answer: True

Solution:

The principal quantum number 'n' determines the size and to a large extent the energy of the orbital.

Correct Answer: False

Solution:

The word 'atom' is derived from the Greek word 'a-tomio', which means 'uncut-able' or 'non-divisible'.

Correct Answer: False

Solution:

The excerpt explains that the word 'atom' is derived from the Greek word 'a-tomio', which means 'uncut-able' or 'non-divisible'.

Correct Answer: True

Solution:

John Dalton, a British school teacher, proposed the atomic theory of matter in 1808, which regarded the atom as the ultimate particle of matter.

Correct Answer: True

Solution:

Dalton's atomic theory successfully explained the law of conservation of mass.

Correct Answer: False

Solution:

The Bohr model could not explain the stability of atoms because it did not account for the emission of radiation by accelerating electrons, leading to the theoretical collapse of the atom.

Correct Answer: False

Solution:

The azimuthal quantum number 'l' defines the shape of the orbital, not its orientation.

Correct Answer: False

Solution:

The early Indian and Greek philosophers speculated about the existence of atoms, but there was no experimental confirmation of their ideas at that time.

Correct Answer: True

Solution:

In hydrogen-like species, the energy of the orbital depends on the principal quantum number 'n'.

Correct Answer: False

Solution:

Dalton's atomic theory could not explain the electrical charging of substances like glass or ebonite when rubbed with silk or fur, which was a known phenomenon at the time.

Correct Answer: False

Solution:

The azimuthal quantum number 'l' defines the three-dimensional shape of the orbital, not its size.

Correct Answer: True

Solution:

The internal structure of atoms is responsible for the diverse chemical behavior of elements.

Correct Answer: True

Solution:

In the Bohr model, the circumference of the electron's orbit is indeed an integral multiple of its de Broglie wavelength.

Correct Answer: True

Solution:

The term 'atom' originates from the Greek word 'a-tomio', which translates to 'uncut-able' or 'non-divisible', reflecting the early belief that atoms were the smallest indivisible components of matter.

Correct Answer: False

Solution:

John Dalton's atomic theory regarded the atom as the ultimate particle of matter, which means it was considered indivisible.

Correct Answer: False

Solution:

Rutherford's model of the atom is similar to a solar system, where electrons move in well-defined orbits around the nucleus, not stationary.

Correct Answer: False

Solution:

Using a thin foil of light atoms like aluminum in Rutherford's experiment would result in different scattering patterns due to the lower atomic mass and charge, affecting the deflection of alpha particles.

Correct Answer: False

Solution:

The photoelectric effect cannot be explained using classical electromagnetic theory; it requires Planck's quantum theory.

Correct Answer: True

Solution:

In hydrogen-like species, the principal quantum number 'n' determines the size and energy of the orbital, with larger 'n' values indicating larger orbitals and higher energy.

Correct Answer: True

Solution:

Dalton's atomic theory successfully explained the law of conservation of mass. However, it could not explain phenomena like electrical charging of substances, which involve the electrical nature of matter.

Correct Answer: True

Solution:

Atoms consist of sub-atomic particles: electrons, protons, and neutrons.

Correct Answer: False

Solution:

The Rutherford model could not explain the stability of atoms because it suggested that electrons would spiral into the nucleus.

Correct Answer: False

Solution:

The word 'atom' is derived from the Greek word 'a-tomio', which means 'uncut-able' or 'non-divisible'.

Correct Answer: True

Solution:

The photoelectric effect is explained by Planck's quantum theory, which describes how light can be considered as discrete packets of energy called quanta or photons.

Correct Answer: False

Solution:

Dalton's atomic theory regarded the atom as the ultimate particle of matter, which means it was considered indivisible.

Correct Answer: True

Solution:

Dalton's atomic theory successfully explained the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.

Correct Answer: True

Solution:

Rutherford's model of the atom is likened to a small-scale solar system, where electrons orbit the nucleus similar to planets orbiting the sun.

Correct Answer: False

Solution:

Early Indian and Greek philosophers believed that atoms were the fundamental building blocks of matter and were indivisible.

Correct Answer: True

Solution:

The term 'atom' comes from the Greek word 'a-tomio', reflecting the belief that atoms were the smallest indivisible units of matter.

Correct Answer: False

Solution:

The atomic theory of matter was first proposed on a firm scientific basis by John Dalton in 1808, not by early Indian and Greek philosophers.

Correct Answer: True

Solution:

Dalton's concept of the atom as an indivisible particle was contradicted by the discovery of sub-atomic particles, which showed that atoms are made of electrons, protons, and neutrons.