STRUCTURE OF ATOM MCQs for NEET UG CHEMISTRY — Easy, Page 3

61 Easy

Remembering +4 | -1

The present atomic weight scale is:

View answer and explanation

Correct option: A

{C^{12}}

Option A is correct because the current atomic weight scale is based on Carbon-12 (

{C^{12}}

).

Solution approach

Correct Answer + Core Reason: ${C^{12}}$ . The present atomic weight scale is based on the ${C^{12}}$ isotope.

Step-by-Step Solution:

Step 1: Recall the standard for atomic weight scale.

Step 2: Identify the element used as the standard.

Step 3: The standard is ${C^{12}}$ .

Why other options are wrong:
- Option B: ${O^{16}}$ was used in the past but is not the current standard.
- Option C: ${H^1}$ is not used as the standard.
- Option D: ${C^{13}}$ is an isotope of carbon but not the standard.

Key Concepts:
- Atomic weight scale
- Standard element

Memory Aids:
- Carbon-12 is the standard.

Quick Tips:
- Remember the current standard for atomic weight.

62 Easy

62 of 176Question permalink

Understanding +4 | -1

Light, a well known form of energy, is treated as a form of matter, by saying that it consist of :

View answer and explanation

Correct option: A

Photons or bundles of energy

Option A (Photons or bundles of energy) is correct because light is considered to be composed of photons, which are discrete packets of energy.

Solution approach

Correct Answer + Core Reason: Photons or bundles of energy. Light is quantized into photons.

Step-by-Step Solution:

Step 1: Recall the dual nature of light.

Step 2: Light exhibits both wave-like and particle-like properties. The particle-like property is described by photons.

Step 3: Define photons.

Step 4: Photons are discrete packets of energy (quanta) that make up electromagnetic radiation, including light.

Step 5: Relate photons to the question.

Step 6: Therefore, light can be treated as a form of matter by considering it to consist of photons or bundles of energy.

Why other options are wrong:
- Option B: Electrons or a wave like matter. While electrons exhibit wave-particle duality, they are not the primary constituents of light. Light itself is the wave.
- Option C: Neutrons, since electrically neutral. Neutrons are subatomic particles found in the nucleus of atoms. They are not related to light.
- Option D: None of the above. Option A is correct, so this option is incorrect.

Key Concepts:
- Wave-particle duality of light
- Photons
- Quantum mechanics

Memory Aids:
- Light = Photons = Energy packets

Quick Tips:
- Remember that light has a dual nature: wave and particle (photon).

63 Easy

63 of 176Question permalink

Application +4 | -1

The electronic configuration of a dipositive ion

{M^{2 + }}

is 2, 8, 14 and its mass number is 56. The number of neutrons present is

View answer and explanation

Correct option: C

30

Option C (30) is correct because the number of neutrons is calculated by subtracting the number of protons from the mass number. The number of protons is found from the electronic configuration.

Solution approach

Correct Answer + Core Reason: 30. The number of neutrons is calculated by subtracting the number of protons from the mass number.

Step-by-Step Solution:

Step 1: The electronic configuration of $M^{2+}$ is 2, 8, 14. This means that the number of electrons in $M^{2+}$ is 2 + 8 + 14 = 24.

Step 2: Since $M^{2+}$ has a +2 charge, it means that the neutral atom M has lost 2 electrons. Therefore, the number of electrons in the neutral atom M is 24 + 2 = 26. The number of protons in the neutral atom M is equal to the number of electrons, which is 26.

Step 3: The mass number of M is 56. The mass number is the sum of the number of protons and the number of neutrons.

Step 4: Number of neutrons = Mass number - Number of protons = 56 - 26 = 30.

Why other options are wrong:
- Option A: Incorrect calculation of neutrons.
- Option B: Incorrect calculation of neutrons.
- Option D: Incorrect calculation of neutrons.

Key Concepts:
- Electronic configuration
- Ions and charge
- Mass number
- Number of protons, neutrons, and electrons

Memory Aids:
- Atomic number = number of protons = number of electrons (in a neutral atom)
- Mass number = number of protons + number of neutrons

Quick Tips:
- Remember to account for the charge when determining the number of electrons in an ion.

64 Easy

64 of 176Question permalink

Understanding +4 | -1

For an electron, if the uncertainty in velocity is ∆v, the uncertainty in its position (∆x) is given by:

View answer and explanation

Correct option: C

\frac{h}{{4\pi m\Delta v}}

Option C (

\frac{h}{{4\pi m\Delta v}}

) is correct because it represents the Heisenberg uncertainty principle, relating uncertainty in position and momentum.

Solution approach

Correct Answer + Core Reason: $\frac{h}{{4\pi m\Delta v}}$ . This is the Heisenberg Uncertainty Principle.

Step-by-Step Solution:

Step 1: State the Heisenberg Uncertainty Principle: The product of the uncertainties in position ( $\Delta x$ ) and momentum ( $\Delta p$ ) is greater than or equal to $\frac{h}{{4\pi }}$ .

Step 2: Write the formula: $\Delta x \Delta p \ge \frac{h}{{4\pi }}$

Step 3: Express momentum in terms of mass and velocity: $\Delta p = m \Delta v$

Step 4: Substitute $\Delta p$ in the uncertainty principle: $\Delta x (m \Delta v) \ge \frac{h}{{4\pi }}$

Step 5: Solve for $\Delta x$ : $\Delta x \ge \frac{h}{{4\pi m \Delta v}}$

Step 6: Final Answer: $\Delta x = \frac{h}{{4\pi m\Delta v}}$

Why other options are wrong:
- Option A: $\frac{h}{2}\pi m\Delta v$ : This is incorrect because it has the wrong constants and places the mass and velocity in the numerator instead of the denominator.
- Option B: $\frac{{2\pi }}{{hm\Delta v}}$ : This is incorrect because it inverts the entire expression and has incorrect constants.
- Option D: $\frac{{2\pi m}}{{h\Delta v}}$ : This is incorrect because it inverts the expression and has the mass in the numerator and Planck's constant in the denominator.

Key Concepts:
- Heisenberg Uncertainty Principle
- Uncertainty in position and momentum

Memory Aids:
- Remember the Heisenberg Uncertainty Principle formula: $\Delta x \Delta p \ge \frac{h}{{4\pi }}$

Quick Tips:
- Always remember the correct form of the Heisenberg Uncertainty Principle.
- Pay attention to the constants and their positions in the formula.

65 Easy

65 of 176Question permalink

Analyzing +4 | -1

Assertion : Total number of orbitals associated with principal quantum number

n=3

is 6 .

Reason : Number of orbitals in a shell equals to

2 n

.

View answer and explanation

Correct option: D

If the assertion and reason both are false.

The assertion is false because the total number of orbitals associated with n=3 is 9, and the reason is also false because the number of orbitals in a shell is n².

Solution approach

Correct Answer + Core Reason: If the assertion and reason both are false. The assertion is incorrect because the total number of orbitals for n=3 is 9, and the reason is incorrect because the number of orbitals in a shell is $n^2$ .

Step-by-Step Solution:

Step 1: Analyze the assertion: Total number of orbitals associated with principal quantum number n=3 is 6.

Step 2: For n=3, the possible values of l are 0, 1, and 2. This corresponds to the 3s, 3p, and 3d subshells.

Step 3: The number of orbitals in each subshell is given by 2l+1. So, for l=0 (s), there is 1 orbital; for l=1 (p), there are 3 orbitals; and for l=2 (d), there are 5 orbitals.

Step 4: The total number of orbitals for n=3 is 1 + 3 + 5 = 9. Therefore, the assertion is false.

Step 5: Analyze the reason: Number of orbitals in a shell equals to 2n.

Step 6: The number of orbitals in a shell is given by $n^2$ . For example, when n=1, there is 1 orbital; when n=2, there are 4 orbitals; and when n=3, there are 9 orbitals. Therefore, the reason is false.

Why other options are wrong:
- Option A: This option is incorrect because both the assertion and the reason are false.
- Option B: This option is incorrect because both the assertion and the reason are false.
- Option C: This option is incorrect because both the assertion and the reason are false.

Key Concepts:
- Principal quantum number (n)
- Number of orbitals in a subshell (2l+1)
- Number of orbitals in a shell ( $n^2$ )

Memory Aids:
- Remember that the number of orbitals in a shell is $n^2$ .

Quick Tips:
- Always calculate the number of orbitals for each subshell and sum them to find the total number of orbitals for a given n.
- Remember the formula for the number of orbitals in a shell: $n^2$ .

66 Easy

66 of 176Question permalink

Understanding +4 | -1

An isotone of

\mathrm{\;}_{32}^{76}\mathrm{Ge}

is

View answer and explanation

Correct option: B

\mathrm{\;}_{33}^{77}\mathrm{As}

Option B (

\mathrm{\;}_{33}^{77}\mathrm{As}

) is correct because isotones have the same number of neutrons.

^{76}_{32}Ge

has 76-32 = 44 neutrons.

^{77}_{33}As

has 77-33 = 44 neutrons.

Solution approach

Correct Answer + Core Reason: $\mathrm{\;}_{33}^{77}\mathrm{As}$ . Isotones have the same number of neutrons.

Step-by-Step Solution:

Step 1: Recall the definition of isotones: Isotones are atoms with the same number of neutrons but different numbers of protons.

Step 2: Calculate the number of neutrons in $\mathrm{\;}_{32}^{76}\mathrm{Ge}$ :
- Number of neutrons = Mass number - Atomic number = 76 - 32 = 44.

Step 3: Calculate the number of neutrons in each option:
- A: $\mathrm{\;}_{32}^{77}\mathrm{Ge}$ : 77 - 32 = 45 neutrons.
- B: $\mathrm{\;}_{33}^{77}\mathrm{As}$ : 77 - 33 = 44 neutrons.
- C: $\mathrm{\;}_{34}^{77}\mathrm{Se}$ : 77 - 34 = 43 neutrons.
- D: $\mathrm{\;}_{36}^{78}\mathrm{Sc}$ : 78 - 36 = 42 neutrons.

Step 4: Identify the option with 44 neutrons.

Why other options are wrong:
- Option A: $\mathrm{\;}_{32}^{77}\mathrm{Ge}$ has 45 neutrons, not 44.
- Option C: $\mathrm{\;}_{34}^{77}\mathrm{Se}$ has 43 neutrons, not 44.
- Option D: $\mathrm{\;}_{36}^{78}\mathrm{Sc}$ has 42 neutrons, not 44.

Key Concepts:
- Isotones
- Atomic number
- Mass number
- Number of neutrons

Memory Aids:
- Isotones have the same 'n'umber of neutrons.

Quick Tips:
- Remember that the number of neutrons = Mass number - Atomic number.

67 Easy

67 of 176Question permalink

Analyzing +4 | -1

Statement (S): Humphrey series discovered in

\mathrm{H}

atomic spectra has lowest energy radiations.

Explanation (E): The series belongs to

n_1=6.

View answer and explanation

Correct option: C

Both S and E are correct and E is correct explanation of S

Option C (Both S and E are correct and E is correct explanation of S) is correct because the Humphrey series, with

n_1 = 6

, involves transitions to a higher energy level, resulting in lower energy radiations.

Solution approach

Correct Answer + Core Reason: Both S and E are correct and E is correct explanation of S. The Humphrey series has the lowest energy radiations because it corresponds to transitions to the $n=6$ level.

Step-by-Step Solution:

Step 1: Statement (S): Humphrey series discovered in $\mathrm{H}$ atomic spectra has lowest energy radiations. The Humphrey series corresponds to transitions where the electron falls to the $n_1 = 6$ energy level.

Step 2: Explanation (E): The series belongs to $n_1 = 6$ . This explains why the Humphrey series has the lowest energy radiations because the energy difference between the levels involved in the transitions is smaller compared to other series (Lyman, Balmer, Paschen, Brackett).

Step 3: Since the transitions involve the $n=6$ level, the energy of the emitted photons is the lowest among all series.

Step 4: Final Answer: Both S and E are correct and E is the correct explanation of S

Why other options are wrong:
- Option A (S is correct but E is wrong): E is correct, as it explains why S is correct.
- Option B (S is wrong but E is correct): S is correct, so this option is incorrect.
- Option D (Both S and E are correct but E is not correct explanation of S): E is the correct explanation for S.

Key Concepts:
- Hydrogen Atomic Spectra
- Spectral Series (Lyman, Balmer, Paschen, Brackett, Humphrey)
- Energy Levels
- Electronic Transitions

Memory Aids:
- Remember the order of the series and their corresponding $n_1$ values: Lyman (1), Balmer (2), Paschen (3), Brackett (4), Humphrey (6).

Quick Tips:
- The series with the highest $n_1$ value will have the lowest energy radiations.

68 Easy

68 of 176Question permalink

Understanding +4 | -1

\left( {_{32}G{e^{76}},\,{\,_{34}}S{e^{76}}} \right)

and

\left( {_{14}S{i^{30}},\,{\,_{16}}{S^{32}}} \right)

are examples of

View answer and explanation

Correct option: B

Isobars and Isotones

Option B (Isobars and Isotones) is correct because

_{32}Ge^{76}

and

_{34}Se^{76}

have the same mass number (isobars), and

_{14}Si^{30}

and

_{16}S^{32}

have the same number of neutrons (isotones).

Solution approach

Correct Answer + Core Reason: Isobars and Isotones. $_{32}Ge^{76}$ and $_{34}Se^{76}$ are isobars, and $_{14}Si^{30}$ and $_{16}S^{32}$ are isotones.

Step-by-Step Solution:

Step 1: Define isotopes, isobars, and isotones.
- Isotopes: Atoms with the same atomic number but different mass numbers.
- Isobars: Atoms with different atomic numbers but the same mass number.
- Isotones: Atoms with different atomic numbers and different mass numbers, but the same number of neutrons.

Step 2: Analyze the first pair: $_{32}Ge^{76}$ and $_{34}Se^{76}$ .
- Atomic numbers are different (32 and 34).
- Mass numbers are the same (76).
- Therefore, they are isobars.

Step 3: Analyze the second pair: $_{14}Si^{30}$ and $_{16}S^{32}$ .
- Atomic numbers are different (14 and 16).
- Mass numbers are different (30 and 32).
- Number of neutrons in $_{14}Si^{30}$ : 30 - 14 = 16.
- Number of neutrons in $_{16}S^{32}$ : 32 - 16 = 16.
- Therefore, they are isotones.

Why other options are wrong:
- Option A: Isotopes and Isobars - The second pair are not isotopes.
- Option C: Isotones and Isotopes - The first pair are not isotones, and the second pair are not isotopes.
- Option D: Isotopes and isotopes - Neither pair are isotopes.

Key Concepts:
- Isotopes
- Isobars
- Isotones

Memory Aids:
- Isotopes: Same proton number.
- Isobars: Same mass number.
- Isotones: Same number of neutrons.

Quick Tips:
- Calculate the number of neutrons (mass number - atomic number) to identify isotones.

69 Easy

69 of 176Question permalink

Remembering +4 | -1

de Broglie equation is a relationship between:

View answer and explanation

Correct option: B

Wavelength of an electron and its momentum

Option B (Wavelength of an electron and its momentum) is correct because the de Broglie equation directly relates wavelength and momentum.

Solution approach

Correct Answer + Core Reason: Wavelength of an electron and its momentum. The de Broglie equation relates the wavelength of a particle to its momentum.

Step-by-Step Solution:

Step 1: Recall the de Broglie equation: $\lambda = \frac{h}{p}$ , where $\lambda$ is the wavelength, $h$ is Planck's constant, and $p$ is the momentum.

Step 2: The equation shows a direct relationship between wavelength and momentum.

Step 3: Therefore, the de Broglie equation is a relationship between the wavelength of an electron and its momentum.

Why other options are wrong:
- Option A: Position of an electron and its momentum: This is related to the Heisenberg Uncertainty Principle, not the de Broglie equation.
- Option C: Mass of an electron and its energy: This is related to mass-energy equivalence ( $E=mc^2$ ), not the de Broglie equation.
- Option D: Wavelength of an electron and its frequency: This is related to the wave equation ( $v = f\lambda$ ) but not directly the de Broglie equation.

Key Concepts:
- de Broglie wavelength
- Momentum

Memory Aids:
- $\lambda = \frac{h}{p}$ (de Broglie equation)

Quick Tips:
- Remember the de Broglie equation and its components.

70 Easy

70 of 176Question permalink

Understanding +4 | -1

The hydrogen spectrum from an incandescent source of hydrogen is:

View answer and explanation

Correct option: B

A line spectrum in emission

Option B (A line spectrum in emission) is correct because an incandescent source of hydrogen emits light at specific wavelengths, resulting in a line spectrum.

Solution approach

Correct Answer + Core Reason: A line spectrum in emission. Incandescent hydrogen emits light at specific wavelengths.

Step-by-Step Solution:

Step 1: The question asks about the type of spectrum produced by an incandescent source of hydrogen.

Step 2: Recall that an incandescent source emits light due to its high temperature. When hydrogen gas is heated, the atoms are excited and emit light at specific wavelengths, creating a line spectrum.

Step 3: Since the hydrogen is emitting light, it is an emission spectrum.

Why other options are wrong:
- Option A: A band spectrum in emission is incorrect because band spectra are typically produced by molecules, not individual atoms like hydrogen.
- Option C: A band spectrum in absorption is incorrect because absorption spectra occur when light passes through a substance, and specific wavelengths are absorbed. This is not the case for an incandescent source.
- Option D: A line spectrum in absorption is incorrect because it is an emission spectrum, not an absorption spectrum.

Key Concepts:
- Emission spectrum
- Absorption spectrum
- Line spectrum
- Band spectrum
- Incandescence

Memory Aids:
- Emission: light is emitted
- Absorption: light is absorbed

Quick Tips:
- Remember the difference between emission and absorption spectra.

71 Easy

71 of 176Question permalink

Remembering +4 | -1

Which is not electromagnetic radiation?

View answer and explanation

Correct option: C

Cathode rays

Option C (Cathode rays) is correct because cathode rays are a stream of electrons, which are matter particles, while the other options are electromagnetic radiation.

Solution approach

Correct Answer + Core Reason: Cathode rays. Cathode rays are a stream of electrons, not electromagnetic radiation.

Step-by-Step Solution:

Step 1: Recall the definition of electromagnetic radiation. Electromagnetic radiation consists of photons and does not require a medium to propagate.

Step 2: Identify the nature of each option:
- Infrared rays: Electromagnetic radiation.
- X-rays: Electromagnetic radiation.
- Cathode rays: Stream of electrons.
- γ-rays: Electromagnetic radiation.

Step 3: Determine which option is not electromagnetic radiation.

Why other options are wrong:
- Option A (Infrared rays): Infrared rays are part of the electromagnetic spectrum.
- Option B (X-rays): X-rays are part of the electromagnetic spectrum.
- Option D (γ-rays): Gamma rays are part of the electromagnetic spectrum.

Key Concepts:
- Electromagnetic radiation
- Cathode rays

Memory Aids:
- Remember that cathode rays are streams of electrons, not photons.

Quick Tips:
- Differentiate between electromagnetic radiation and particle beams.

72 Easy

72 of 176Question permalink

Remembering +4 | -1

Line spectra is characteristic of :

View answer and explanation

Correct option: A

Atoms

Option A (Atoms) is correct because line spectra are produced by the transitions of electrons within individual atoms.

Solution approach

Correct Answer + Core Reason: Atoms. Line spectra are characteristic of electronic transitions within atoms.

Step-by-Step Solution:

Step 1: Recall the origin of line spectra. Line spectra are produced when electrons transition between energy levels within an atom.

Step 2: These transitions emit photons of specific wavelengths, resulting in discrete lines in the spectrum.

Step 3: Molecules, radicals, and ions can also have spectra, but they are typically more complex (band spectra) due to vibrational and rotational modes.

Step 4: Line spectra are most characteristic of individual atoms.

Why other options are wrong:
- Option B: Molecules: Molecules typically exhibit band spectra due to vibrational and rotational energy levels in addition to electronic levels.
- Option C: Radicals: Radicals, like molecules, can have more complex spectra than simple line spectra.
- Option D: Ions: Ions can also exhibit spectra, but line spectra are most fundamentally associated with atoms.

Key Concepts:
- Line spectra
- Atomic spectra
- Electronic transitions
- Energy levels

Memory Aids:
- Remember that line spectra are associated with electronic transitions in atoms.

Quick Tips:
- Understand the relationship between electronic transitions and the wavelengths of emitted light.

73 Easy

73 of 176Question permalink

Understanding +4 | -1

Two electron in an atm of an element cannot have:

View answer and explanation

Correct option: D

An identical set of quantum numbers

The correct answer is D (An identical set of quantum numbers) because, according to the Pauli Exclusion Principle, no two electrons in an atom can have the same set of all four quantum numbers.

Solution approach

Correct Answer + Core Reason: An identical set of quantum numbers. The Pauli Exclusion Principle states that no two electrons in an atom can have the same set of quantum numbers.

Step-by-Step Solution:

Step 1: State the Pauli Exclusion Principle: No two electrons in an atom can have the same set of all four quantum numbers (n, l, ml, ms).

Step 2: Explain the implication of the Pauli Exclusion Principle: This means that if two electrons have the same n, l, and ml values, they must have different ms values (i.e., opposite spins).

Step 3: Conclusion: Therefore, two electrons in an atom cannot have an identical set of quantum numbers.

Why other options are wrong:
- Option A (The same principle quantum number): Electrons can have the same principal quantum number (n), which indicates the energy level or shell.
- Option B (The same azimuthal quantum number): Electrons can have the same azimuthal quantum number (l), which indicates the shape of the orbital (s, p, d, f).
- Option C (The same magnetic quantum number): Electrons can have the same magnetic quantum number (ml), which indicates the orientation of the orbital in space.

Key Concepts:
- Pauli Exclusion Principle
- Quantum Numbers (n, l, ml, ms)
- Atomic Structure

Memory Aids:
- Remember that the Pauli Exclusion Principle is about *all four* quantum numbers being different.

Quick Tips:
- Understand the meaning of each quantum number and how they relate to the electron's state in an atom.

74 Easy

74 of 176Question permalink

Analyzing +4 | -1

Which set of quantum numbers are not possible

View answer and explanation

Correct option: B

\begin{array}{cccccccccccccccccccc} n&l&m&s\\ 2&2&1&{ + 1/2} \end{array}

Option B is not possible because if n=2, l can only be 0 or 1. l cannot be equal to n.

Solution approach

Correct Answer + Core Reason: $\begin{array}{cccccccccccccccccccc}n&l&m&s\\2&2&1&{ + 1/2}\end{array}$ . This set of quantum numbers is not possible because the value of $l$ must be less than $n$ .

Step-by-Step Solution:

Step 1: Recall the rules for quantum numbers:
- Principal quantum number (n): n = 1, 2, 3, ...
- Azimuthal quantum number (l): l = 0, 1, 2, ..., (n-1)
- Magnetic quantum number (m): m = -l, -l+1, ..., 0, ..., l-1, l
- Spin quantum number (s): s = +1/2 or -1/2

Step 2: Analyze option A: n=3, l=2, m=0, s=+1/2. Since l < n and -l <= m <= l, this set is possible.

Step 3: Analyze option B: n=2, l=2, m=1, s=+1/2. Since l must be less than n, l cannot be 2 when n is 2. This set is NOT possible.

Step 4: Analyze option C: n=1, l=0, m=0, s=-1/2. Since l < n and -l <= m <= l, this set is possible.

Step 5: Analyze option D: n=3, l=2, m=-2, s=+1/2. Since l < n and -l <= m <= l, this set is possible.

Why other options are wrong:
- Option A: This set of quantum numbers is valid because l (2) is less than n (3), and m (0) is within the range -l to +l.
- Option C: This set of quantum numbers is valid because l (0) is less than n (1), and m (0) is within the range -l to +l.
- Option D: This set of quantum numbers is valid because l (2) is less than n (3), and m (-2) is within the range -l to +l.

Key Concepts:
- Quantum numbers (n, l, m, s)
- Rules for allowed values of quantum numbers

Memory Aids:
- Remember the ranges for each quantum number: n > l, -l <= m <= l, s = +/- 1/2

Quick Tips:
- Always check if the value of l is less than n.
- Ensure that m is within the range of -l to +l.

75 Easy

75 of 176Question permalink

Understanding +4 | -1

The de Broglie wavelength associated with a material particle is:

View answer and explanation

Correct option: A

Inversely proportional to momentum

Option A (3) is correct because the M-shell corresponds to n=3, and for each n, there are n possible values of l (0 to n-1), representing the number of elliptical orbits, including the circular one.

Solution approach

Correct Answer + Core Reason: 3. The number of elliptical orbits, including circular orbits, in the M-shell is equal to the principal quantum number n=3.

Step-by-Step Solution:

Step 1: Identify the principal quantum number (n) for the M-shell. The M-shell corresponds to n = 3.

Step 2: Recall that for a given principal quantum number n, the azimuthal quantum number (l) can have values from 0 to n-1.

Step 3: Determine the possible values of l for n = 3:
- l = 0 (s orbital, spherical)
- l = 1 (p orbital, dumbbell-shaped)
- l = 2 (d orbital, more complex shape)

Step 4: The number of elliptical orbits, including the circular orbit, is equal to the number of possible values of l, which is 3.

Why other options are wrong:
- Option B: 4: This is incorrect because it corresponds to the N-shell (n=4).
- Option C: 2: This is incorrect because it corresponds to the L-shell (n=2).
- Option D: 1: This is incorrect because it corresponds to the K-shell (n=1).

Key Concepts:
- Principal quantum number (n)
- Azimuthal quantum number (l)
- Electronic shells (K, L, M, N...)
- Atomic orbitals (s, p, d, f...)

Memory Aids:
- Remember that the M-shell corresponds to n=3.
- For each n, there are n possible values of l.

Quick Tips:
- The number of orbitals in a shell is $n^2$ , and the number of electrons is $2n^2$ .

76 Easy

76 of 176Question permalink

Understanding +4 | -1

Which best describe the emission spectra of atomic hydrogen?

View answer and explanation

Correct option: C

Several discrete series of lines with both intensity and spacings between lines decreasing as the wave number increase within each series

Option C (Several discrete series of lines with both intensity and spacings between lines decreasing as the wave number increase within each series) is correct because it accurately describes the characteristics of the hydrogen emission spectrum.

Solution approach

Correct Answer + Core Reason: Several discrete series of lines with both intensity and spacings between lines decreasing as the wave number increase within each series. This accurately describes the hydrogen emission spectrum.

Step-by-Step Solution:

Step 1: The question asks for the best description of the emission spectra of atomic hydrogen.

Step 2: Recall the characteristics of the hydrogen emission spectrum:
- It consists of several discrete series of lines (Lyman, Balmer, Paschen, etc.).
- Within each series, the intensity of the lines decreases as the wave number increases.
- The spacing between the lines also decreases as the wave number increases.

Step 3: Therefore, the best description is "Several discrete series of lines with both intensity and spacings between lines decreasing as the wave number increase within each series".

Why other options are wrong:
- Option A: A series of only four lines is incorrect because there are more than four lines in the hydrogen emission spectrum.
- Option B: A discrete series of lines of equal intensity and equally spaced with respect to wavelength is incorrect because the lines have different intensities and are not equally spaced.
- Option D: A continuous emission of radiation of all frequencies is incorrect because the hydrogen emission spectrum is discrete, not continuous.

Key Concepts:
- Hydrogen emission spectrum
- Discrete spectrum
- Intensity of spectral lines
- Wave number

Memory Aids:
- Discrete lines, not continuous.

Quick Tips:
- Remember the general characteristics of the hydrogen emission spectrum.

77 Easy

77 of 176Question permalink

Analyzing +4 | -1

Assertion: Nuclear binding energy per nucleon is in the order

{ }_4^9 {Be}>{ }_3^7 {Li}>{ }_2^4 {He}

.

Reason: Binding energy per nucleon increases linearly with difference in number of neutrons and protons.

View answer and explanation

Correct option: D

If both the Assertion and Reason are incorrect.

Option D is correct because both the assertion and the reason are incorrect regarding nuclear binding energy and its relationship with nucleon numbers.

Solution approach

Correct Answer + Core Reason: If both the Assertion and Reason are incorrect. The binding energy per nucleon does not follow the order given in the assertion, and the reason is incorrect as it doesn't increase linearly with the difference in the number of neutrons and protons.

Step-by-Step Solution:

Step 1: Analyze the assertion. The binding energy per nucleon generally increases with mass number up to iron (Fe) and then gradually decreases. The order given is incorrect.

Step 2: Analyze the reason. The binding energy per nucleon is related to the strong nuclear force and the stability of the nucleus, but it doesn't increase linearly with the difference in the number of neutrons and protons. The relationship is more complex and depends on various factors.

Step 3: Conclude that both the assertion and the reason are incorrect.

Why other options are wrong:
- Option A: The assertion and reason are not both correct, so this option is wrong.
- Option B: The assertion and reason are not both correct, so this option is wrong.
- Option C: The assertion is incorrect, so this option is wrong.

Key Concepts:
- Nuclear binding energy
- Binding energy per nucleon
- Nuclear stability

Memory Aids:
- Binding energy per nucleon curve

Quick Tips:
- Remember the general trend of binding energy per nucleon with mass number.

78 Easy

78 of 176Question permalink

Analyzing +4 | -1

Which has minimum number of unpaired d-electrons?

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{o}^{2 + }}

Option C (

\mathrm{C}{\mathrm{o}^{2 + }}

) is correct because it has the fewest unpaired d-electrons compared to the other options.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{o}^{2 + }}$ . Cobalt(II) has the minimum number of unpaired d-electrons among the given options.

Step-by-Step Solution:

Step 1: Write the electronic configuration of each ion.
- Fe3+ (Z=26-3=23): [Ar] 3d5
- Co3+ (Z=27-3=24): [Ar] 3d6
- Co2+ (Z=27-2=25): [Ar] 3d7
- Mn2+ (Z=25-2=23): [Ar] 3d5

Step 2: Determine the number of unpaired d-electrons in each ion.
- Fe3+: 3d5 has 5 unpaired electrons.
- Co3+: 3d6 has 4 unpaired electrons (using Hund's rule).
- Co2+: 3d7 has 3 unpaired electrons (using Hund's rule).
- Mn2+: 3d5 has 5 unpaired electrons.

Step 3: Identify the ion with the minimum number of unpaired d-electrons.\nCo2+ has the minimum number of unpaired d-electrons (3).

Why other options are wrong:
- Option A: $\mathrm{F}{\mathrm{e}^{3 + }}$ : Fe3+ has 5 unpaired d-electrons.
- Option B: $\mathrm{C}{\mathrm{o}^{3 + }}$ : Co3+ has 4 unpaired d-electrons.
- Option D: $\mathrm{M}{\mathrm{n}^{2 + }}$ : Mn2+ has 5 unpaired d-electrons.

Key Concepts:
- Electronic configuration
- Hund's rule
- Unpaired electrons

Memory Aids:
- Remember Hund's rule: electrons will individually occupy each orbital within a subshell before pairing.

Quick Tips:
- Write the electronic configuration first.
- Apply Hund's rule to determine the number of unpaired electrons.

79 Easy

79 of 176Question permalink

Application +4 | -1

The mass of an electron is m, its charge e and it is accelerated from rest through a potential difference V. The kinetic energy of the electron in joules will be :

View answer and explanation

Correct option: B

eV

Option B (eV) is correct because the kinetic energy gained by an electron accelerated through a potential difference V is equal to the charge of the electron (e) multiplied by the potential difference (V).

Solution approach

Correct Answer + Core Reason: eV. The kinetic energy of an electron accelerated through a potential difference V is given by eV.

Step-by-Step Solution:

Step 1: Recall the relationship between potential difference and kinetic energy.

Step 2: The potential difference (V) is defined as the work done per unit charge to move a charge between two points.

Step 3: The work done on the electron is equal to the change in its kinetic energy.

Step 4: Therefore, the kinetic energy (KE) gained by the electron is equal to the charge of the electron (e) multiplied by the potential difference (V): $KE = eV$ .

Why other options are wrong:
- Option A: V is incorrect because it doesn't account for the charge of the electron.
- Option C: MeV is incorrect because it implies a much larger energy scale (Mega electron volts).
- Option D: None of these is incorrect because eV is the correct answer.

Key Concepts:
- Potential difference
- Kinetic energy
- Work-energy theorem

Memory Aids:
- Remember that the kinetic energy gained by a charge q accelerated through a potential difference V is qV.

Quick Tips:
- Pay attention to the units. In this case, the kinetic energy is in electron volts (eV), which can be converted to joules by multiplying by the elementary charge (1.602 × 10^-19 C).

80 Easy

80 of 176Question permalink

Remembering +4 | -1

Non-directional orbital is

View answer and explanation

Correct option: D

3s

Option D is correct because s-orbitals are spherically symmetrical and thus non-directional, unlike p and d orbitals which have specific directional properties.

Solution approach

Correct Answer + Core Reason: 3s. s-orbitals are spherically symmetrical and non-directional.

Step-by-Step Solution:

Step 1: Recall the shapes of atomic orbitals.
- s-orbitals are spherically symmetrical and have no directional preference.
- p-orbitals have a dumbbell shape and are directional, oriented along the x, y, or z axes.
- d-orbitals have more complex shapes and are also directional.
- f-orbitals have even more complex shapes and are also directional.

Step 2: Identify the non-directional orbital.
- The 3s orbital is an s-orbital, which is spherically symmetrical and non-directional.

Why other options are wrong:
- Option A: 4p is directional.
- Option B: 4d is directional.
- Option C: 4f is directional.

Key Concepts:
- Atomic orbitals (s, p, d, f)
- Directional properties of orbitals

Memory Aids:
- Remember that s-orbitals are spherically symmetrical and non-directional.

Quick Tips:
- Visualize the shapes of the orbitals to determine their directional properties.

81 Easy

81 of 176Question permalink

Application +4 | -1

The de-Broglie wavelength associated with a particle of mass

10^{-6} {~kg}

moving with a velocity of

10 {~ms}^{-1}

is :

View answer and explanation

Correct option: D

6.63 \times 10^{-29} {~m}

The correct answer is D (

6.63 \times 10^{-29} {~m}

) because it is calculated by using the de Broglie wavelength formula:

\lambda = \frac{h}{mv}

.

Solution approach

Correct Answer + Core Reason: $6.63 \times 10^{-29} {~m}$ . The de-Broglie wavelength is calculated using the formula $\lambda = \frac{h}{mv}$ .

Step-by-Step Solution:

Step 1: Identify the given information: mass (m) = $10^{-6} {~kg}$ , velocity (v) = $10 {~ms}^{-1}$ , Planck's constant (h) = $6.63 \times 10^{-34} {~Js}$ .

Step 2: Use the de-Broglie wavelength formula: $\lambda = \frac{h}{mv}$

Step 3: Substitute the given values: $\lambda = \frac{6.63 \times 10^{-34} {~Js}}{(10^{-6} {~kg}) \times (10 {~ms}^{-1})}$

Step 4: Calculate the wavelength: $\lambda = \frac{6.63 \times 10^{-34}}{10^{-5}} {~m}$

Step 5: Simplify: $\lambda = 6.63 \times 10^{-29} {~m}$

Why other options are wrong:
- **Option A ( $6.63 \times 10^{-7} {~m}$ ): This is incorrect. It's likely a result of not correctly substituting the values or making a calculation error with the exponents.
- Option B ( $6.63 \times 10^{-16} {~m}$ ): This is incorrect. It's likely a result of not correctly substituting the values or making a calculation error with the exponents.
- Option C ( $6.63 \times 10^{-21} {~m}$ ): This is incorrect. It's likely a result of not correctly substituting the values or making a calculation error with the exponents.
- Option D ( $6.63 \times 10^{-29} {~m}$ ): This is the correct answer, as shown above.

Key Concepts:
- de-Broglie wavelength
- Planck's constant

Memory Aids:**
- Remember the de-Broglie equation: $\lambda = \frac{h}{mv}$

Quick Tips:
- Pay close attention to the units and exponents when substituting values into the formula.

82 Easy

82 of 176Question permalink

Remembering +4 | -1

Wave nature of electrons was demonstrated by

View answer and explanation

Correct option: B

De-Broglie

Option B (De-Broglie) is correct because De-Broglie proposed the wave nature of electrons.

Solution approach

Correct Answer + Core Reason: De-Broglie. De-Broglie proposed the wave nature of electrons.

Step-by-Step Solution:

Step 1: Recall the historical experiments and theories related to the wave nature of electrons.

Step 2: De Broglie proposed that matter (including electrons) has wave-like properties.

Step 3: Davisson and Germer experimentally confirmed the wave nature of electrons through diffraction experiments.

Why other options are wrong:
- Option A: Schrodinger: Schrodinger developed the wave equation to describe the behavior of quantum mechanical systems, but De Broglie first proposed the wave nature of electrons.
- Option C: Davisson and Garmer: Davisson and Germer experimentally verified De Broglie's hypothesis, but they didn't propose it.
- Option D: Heisenberg: Heisenberg is known for the Uncertainty Principle, not the wave nature of electrons.

Key Concepts:
- Wave-particle duality
- De Broglie hypothesis
- Davisson-Germer experiment

Memory Aids:
- De Broglie: "matter waves"

Quick Tips:
- Remember the contributions of each scientist to the understanding of quantum mechanics.

83 Easy

83 of 176Question permalink

Understanding +4 | -1

The difference between ions and atoms is of:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because ions differ from atoms in size, electron configuration, and the presence of a net charge.

Solution approach

Correct Answer + Core Reason: All of these. Ions and atoms differ in size, configuration, and charge.

Step-by-Step Solution:

Step 1: Define ions and atoms.
- Atoms are neutral species with equal numbers of protons and electrons.
- Ions are charged species formed by the gain or loss of electrons.

Step 2: Explain the differences between ions and atoms:
- Relative size: Ions can be larger (anions) or smaller (cations) than their parent atoms.
- Configuration: The electron configuration changes when an atom gains or loses electrons.
- Presence of charge: Ions have a net positive or negative charge, while atoms are neutral.

Step 3: Analyze the options.
- All the listed differences are valid.

Why other options are wrong:
- Option A: Relative size is a difference, but not the only one.
- Option B: Configuration is a difference, but not the only one.
- Option C: Presence of charge is a difference, but not the only one.

Key Concepts:
- Atoms and ions
- Electron configuration
- Ionic charge
- Atomic and ionic size

Memory Aids:
- Cations are smaller than their parent atoms (positive ions).
- Anions are larger than their parent atoms (negative ions).

Quick Tips:
- The formation of ions affects both the size and the electron configuration of the species.

84 Easy

84 of 176Question permalink

Understanding +4 | -1

In order to designate an orbital n in an atom, the number of quantum number required are:

View answer and explanation

Correct option: C

3

Option C (3) is correct because to designate an orbital in an atom, we need the principal quantum number (n), the azimuthal quantum number (l), and the magnetic quantum number (ml).

Solution approach

Correct Answer + Core Reason: 3. Three quantum numbers (n, l, ml) are required to designate an orbital.

Step-by-Step Solution:

Step 1: Identify the quantum numbers needed to define an orbital.

Step 2: The three quantum numbers required are:
- Principal quantum number (n): Defines the energy level.
- Azimuthal quantum number (l): Defines the shape of the orbital.
- Magnetic quantum number (ml): Defines the orientation of the orbital in space.

Why other options are wrong:
- Option A: 1. One quantum number is not sufficient to define an orbital.
- Option B: 2. Two quantum numbers are not sufficient to define an orbital.
- Option D: 4. Four quantum numbers define an electron, not just the orbital. The fourth quantum number is the spin quantum number (s).

Key Concepts:
- Quantum Numbers
- Atomic Orbitals

Memory Aids:
- n, l, ml define an orbital.

Quick Tips:
- Remember that the spin quantum number defines the electron's spin, not the orbital itself.

85 Easy

85 of 176Question permalink

Remembering +4 | -1

A node is a surface on which the probability of finding an electron is:

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because a node is defined as a region in space where the probability of finding an electron is zero.

Solution approach

Correct Answer + Core Reason: Zero. A node is a region where the probability of finding an electron is zero.

Step-by-Step Solution:

Step 1: Define a node.

Step 2: A node is a point, line, or surface where the wave function ( $\psi$ ) and, consequently, the probability of finding an electron ( $|\psi|^2$ ) is zero.

Step 3: Therefore, the probability of finding an electron at a node is zero.

Why other options are wrong:
- Option B: > 1: Probability cannot be greater than 1.
- Option C: > 10: Probability cannot be greater than 1.
- Option D: > 90: Probability cannot be greater than 1.

Key Concepts:
- Nodes
- Wave function ( $\psi$ )
- Probability density ( $|\psi|^2$ )

Memory Aids:
- Remember that nodes are regions of zero electron density.

Quick Tips:
- Understand the relationship between nodes and the shapes of atomic orbitals.

86 Easy

86 of 176Question permalink

Understanding +4 | -1

For which species, Bohr's theory does not apply:

View answer and explanation

Correct option: B

Be

Option B (Be) is correct because Bohr's theory is only applicable to single-electron species, and Be has 4 electrons.

Solution approach

Correct Answer + Core Reason: Be. Bohr's theory is only applicable to single-electron species.

Step-by-Step Solution:

Step 1: The question asks for which species Bohr's theory does not apply.

Step 2: Recall that Bohr's theory is only applicable to single-electron species (hydrogen-like atoms or ions).

Step 3: Determine the number of electrons in each species:
- H: 1 electron
- Be: 4 electrons
- $\mathrm{H}{\mathrm{e}^ + }$ : 1 electron
- $\mathrm{L}{\mathrm{i}^{2 + }}$ : 1 electron

Step 4: Since Be has more than one electron, Bohr's theory does not apply to it.

Why other options are wrong:
- Option A: H is incorrect because it has only one electron, so Bohr's theory applies.
- Option C: $\mathrm{H}{\mathrm{e}^ + }$ is incorrect because it has only one electron, so Bohr's theory applies.
- Option D: $\mathrm{L}{\mathrm{i}^{2 + }}$ is incorrect because it has only one electron, so Bohr's theory applies.

Key Concepts:
- Bohr's theory
- Single-electron species

Memory Aids:
- Bohr's theory only works for hydrogen-like atoms.

Quick Tips:
- Remember that Bohr's theory is limited to single-electron systems.

87 Easy

87 of 176Question permalink

Remembering +4 | -1

The photoelectric effect occurs only when the incident light has more frequency than a certain minimum:

View answer and explanation

Correct option: A

Frequency

Option A (Frequency) is correct because the photoelectric effect requires the incident light to have a frequency above the threshold frequency for the given material.

Solution approach

Correct Answer + Core Reason: Frequency. The photoelectric effect occurs only when the incident light has a frequency greater than the threshold frequency.

Step-by-Step Solution:

Step 1: Define the photoelectric effect. The photoelectric effect is the emission of electrons when light hits a material.

Step 2: Recall the key concept. The energy of a photon is given by $E = hf$ , where $h$ is Planck's constant and $f$ is the frequency. For the photoelectric effect to occur, the photon energy must be greater than the work function of the material.

Step 3: Explain the role of frequency. If the frequency is below a certain minimum (threshold frequency), no electrons will be emitted, regardless of the intensity of the light.

Step 4: Explain why wavelength is not the primary factor. While wavelength is related to frequency ( $c = \lambda f$ ), frequency is the direct determinant of photon energy.

Why other options are wrong:
- Option B: Wavelength. While related to frequency, wavelength is inversely proportional. A shorter wavelength corresponds to a higher frequency, but frequency is the direct cause of the photoelectric effect.
- Option C: Speed. The speed of light is constant and does not determine whether the photoelectric effect will occur.
- Option D: Charge. Charge is not directly related to the photoelectric effect. The photoelectric effect is about energy transfer from photons to electrons.

Key Concepts:
- Photoelectric effect
- Photon energy ( $E=hf$ )
- Threshold frequency
- Work function

Memory Aids:
- Remember that energy is directly proportional to frequency: E = hf

Quick Tips:
- The photoelectric effect is all about the energy of individual photons, which depends on their frequency.

88 Easy

88 of 176Question permalink

Application +4 | -1

Which of the following set of quantum numbers represents the highest energy orbital in a multi electron atom ?

View answer and explanation

Correct option: D

n = 4,l = 1,m = 0,s = - \frac{1}{2}

Option D (

n = 4,l = 1,m = 0,s = - \frac{1}{2}

) is correct because the energy of an orbital in a multi-electron atom primarily depends on the sum of n+l, and when n+l is the same, the orbital with the higher n value has higher energy.

Solution approach

Correct Answer + Core Reason: $n = 4,l = 1,m = 0,s = - \frac{1}{2}$ . Higher n+l value corresponds to higher energy in a multi-electron atom.

Step-by-Step Solution:

Step 1: Recall the (n+l) rule for determining the energy of an orbital in a multi-electron atom.

Step 2: The energy of an orbital is primarily determined by the sum of the principal quantum number (n) and the azimuthal quantum number (l).

Step 3: The higher the (n+l) value, the higher the energy of the orbital. If two orbitals have the same (n+l) value, the orbital with the higher n value has higher energy.

Step 4: Calculate (n+l) for each option:
- A: n=3, l=0, n+l = 3
- B: n=4, l=0, n+l = 4
- C: n=3, l=1, n+l = 4
- D: n=4, l=1, n+l = 5

Step 5: Option D has the highest (n+l) value (5), so it represents the highest energy orbital.

Why other options are wrong:
- Option A: n=3, l=0, n+l = 3, which is the lowest energy among the given options.
- Option B: n=4, l=0, n+l = 4, which is lower energy than option D.
- Option C: n=3, l=1, n+l = 4, which is lower energy than option D.

Key Concepts:
- Quantum numbers (n, l, m, s)
- (n+l) rule
- Energy levels in multi-electron atoms

Memory Aids:
- Remember the (n+l) rule for determining the energy of orbitals.

Quick Tips:
- Calculate (n+l) for each orbital and compare the values to determine the relative energies.

89 Easy

89 of 176Question permalink

Application +4 | -1

In a Hydrogen atom, if the energy of electron in the ground state is

- x\,\,ev,

then that in the 2nd excited state of

H{e^ + }

is

View answer and explanation

Correct option: B

- \frac{4}{9}x\,\,ev

Option B (

- \frac{4}{9}x\\,\,ev

) is correct because the energy is inversely proportional to n squared and directly proportional to Z squared.

Solution approach

Correct Answer + Core Reason: $- \frac{4}{9}x\\,\,ev$ . The energy of an electron is proportional to $Z^2/n^2$ .

Step-by-Step Solution:

Step 1: Define the energy of the electron in the ground state of hydrogen as $E_H = -x$ eV. For hydrogen, $Z=1$ and $n=1$ , so $E_H = -x = -13.6$ eV.

Step 2: The energy of an electron in a hydrogen-like atom (like $He^+$ ) is given by the formula: $E = -13.6 \frac{Z^2}{n^2}$ eV, where $Z$ is the atomic number and $n$ is the principal quantum number.

Step 3: For $He^+$ , $Z=2$ . The 2nd excited state corresponds to $n=3$ (ground state is n=1, 1st excited state is n=2, 2nd excited state is n=3).

Step 4: Calculate the energy of the electron in the 2nd excited state of $He^+$ : $E_{He^+} = -13.6 \frac{2^2}{3^2} = -13.6 \frac{4}{9}$ eV.

Step 5: Since $E_H = -x = -13.6$ eV, substitute $-x$ for $-13.6$ eV: $E_{He^+} = -x \frac{4}{9} = -\frac{4}{9}x$ eV.

Why other options are wrong:
- Option A: $- x\\,\,ev$ : This is the energy of the electron in the ground state of hydrogen, not the 2nd excited state of $He^+$ .
- Option C: $+ 2x\\,\,ev$ : The energy is always negative for bound states, and this option also doesn't account for the correct $n$ and $Z$ values.
- Option D: $- \frac{{9x}}{4}\\,\,ev$ : This is the inverse of the correct ratio of $Z^2/n^2$ .

Key Concepts:
- Bohr's model of the atom
- Energy levels in hydrogen-like atoms
- Atomic number (Z)
- Principal quantum number (n)

Memory Aids:
- Remember the energy formula: $E = -13.6 \frac{Z^2}{n^2}$ eV

Quick Tips:
- Always identify the correct values of $Z$ and $n$ for the given ion and energy level.

90 Easy

90 of 176Question permalink

Application +4 | -1

The value of Planck’s constant is

6.63 \times {10^{ - 34}}\;\mathrm{Js}.

. The velocity of light is

3.0 \times {10^8}\mathrm{\;m\;}{\mathrm{s}^{ - 1}}.

. Which value is closest to the wavelength in nanometre of a quantum of light with frequency of

8 \times {10^{15}}\mathrm{\;}{\mathrm{s}^{ - 1}}?

View answer and explanation

Correct option: C

4 \times {10^{ - 8}}

Option C is correct. Using the formula λ = c/f, the calculated wavelength is approximately 4 × 10^-8 meters, which is 40 nm.

Solution approach

Correct Answer + Core Reason: $4 \times {10^{ - 8}}$ . The wavelength is calculated using the formula $\lambda = c/f$ .

Step-by-Step Solution:

Step 1: Recall the relationship between wavelength, frequency, and speed of light: $\lambda = \frac{c}{f}$ , where $\lambda$ is the wavelength, $c$ is the speed of light, and $f$ is the frequency.

Step 2: Substitute the given values: $\lambda = \frac{3.0 \times 10^8 \text{ m/s}}{8 \times 10^{15} \text{ s}^{-1}}$ .

Step 3: Calculate the wavelength: $\lambda = 0.375 \times 10^{-7} \text{ m} = 3.75 \times 10^{-8} \text{ m}$ .

Step 4: Convert to nanometers: $\lambda = 3.75 \times 10^{-8} \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}} = 37.5 \text{ nm} \approx 4 \times 10^{-8} \text{ m}$ .

Why other options are wrong:
- Option A: $2 \times {10^{ - 25}}$ : Incorrect; this is a completely different order of magnitude and likely a calculation error.
- Option B: $5 \times {10^{ - 18}}$ : Incorrect; this is a completely different order of magnitude and likely a calculation error.
- Option D: $3 \times {10^7}$ : Incorrect; this is a completely different order of magnitude and likely a calculation error. This would be a very long wavelength, not typical for light.

Key Concepts:
- Wavelength
- Frequency
- Speed of light

Memory Aids:
- Remember the formula $\lambda = \frac{c}{f}$ .

Quick Tips:
- Ensure units are consistent before calculating. Convert to nanometers at the end if needed.

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