Chapter 2: Structure of Atom Application Questions | chemistry Class 11 CBSE

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Chapter 2: Structure of Atom Application Questions | chemistry Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A sodium lamp emitting yellow light at a wavelength of 589 nm is used in a street light. Calculate the frequency and wavenumber of this radiation and identify the region of the electromagnetic spectrum to which it belongs.

Using c = νλ: ν = c/λ = (3.0 × 10^8 m/s)/(589 × 10^-9 m) = 5.09 × 10^14 Hz; this falls in the visible region of the electromagnetic spectrum (400–750 nm range).
Wavenumber ν̄ = 1/λ = 1/(589 × 10^-9 m) = 1.698 × 10^6 m^-1 = 1.698 × 10^4 cm^-1.
The energy of one photon of this radiation: E = hν = (6.626 × 10^-34 J·s)(5.09 × 10^14 s^-1) = 3.37 × 10^-19 J.
Since the wavelength (589 nm) lies between 400 nm (violet) and 750 nm (red), this radiation is in the visible portion of the electromagnetic spectrum, specifically the yellow-orange region perceivable by the human eye.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

In a laboratory, students observe that when radiation of wavelength 300 nm strikes a sodium surface, electrons are emitted. The work function of sodium is 2.31 eV. Calculate whether photoelectric emission will occur, and find the kinetic energy of the emitted electrons.

Energy of incident photon: E = hc/λ = (6.626 × 10^-34 × 3.0 × 10^8)/(300 × 10^-9) = 6.626 × 10^-19 J = 6.626 × 10^-19/1.602 × 10^-19 eV = 4.14 eV.
Work function W₀ = 2.31 eV; since incident photon energy (4.14 eV) > work function (2.31 eV), the photoelectric effect will indeed occur.
Kinetic energy of emitted electron = E − W₀ = 4.14 − 2.31 = 1.83 eV = 1.83 × 1.602 × 10^-19 J = 2.93 × 10^-19 J.
This scenario illustrates Einstein's equation KE = hν − hν₀: only the excess energy beyond the work function appears as kinetic energy of the ejected electron; higher incident frequency gives more energetic photoelectrons.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A hydrogen atom in an excited state emits light when the electron falls from n = 4 to n = 2. Identify the spectral series this transition belongs to, and calculate the energy and frequency of the emitted photon.

Since the transition ends at n = 2, this belongs to the Balmer series, which is the only hydrogen spectral series in the visible region of the electromagnetic spectrum.
ΔE = RH(1/nf² − 1/ni²) = 2.18 × 10^-18(1/4 − 1/16) = 2.18 × 10^-18 × (4−1)/16 = 2.18 × 10^-18 × 3/16 = 4.09 × 10^-19 J.
Frequency: ν = ΔE/h = (4.09 × 10^-19)/(6.626 × 10^-34) = 6.17 × 10^14 Hz; wavelength λ = c/ν = (3.0 × 10^8)/(6.17 × 10^14) ≈ 486 nm (blue-green region of visible light).
This transition produces the H_β line of the Balmer series; in practice, hydrogen discharge tubes show this as a distinct blue-green spectral line useful in spectroscopic identification of hydrogen.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

With reference to the de Broglie hypothesis, compare the wavelengths of a golf ball (mass = 0.045 kg, velocity = 40 m/s) and an electron moving at 2 × 10^6 m/s. Comment on the significance of each wavelength.

De Broglie wavelength for golf ball: λ = h/mv = (6.626 × 10^-34)/(0.045 × 40) = (6.626 × 10^-34)/1.8 = 3.68 × 10^-34 m. This is unimaginably small — far smaller than any atomic dimension — making wave properties of the golf ball completely undetectable.
Wavelength for electron: λ = (6.626 × 10^-34)/(9.11 × 10^-31 × 2 × 10^6) = (6.626 × 10^-34)/(1.822 × 10^-24) = 3.64 × 10^-10 m = 364 pm. This is comparable to atomic and interatomic distances.
The electron's wavelength (~364 pm) is detectable and comparable to X-ray wavelengths used in crystallography, explaining why electrons can be diffracted by crystal lattices — experimental proof of matter waves.
This comparison illustrates that wave behaviour is only significant for sub-atomic particles (small mass); for macroscopic objects, wave character is negligible and classical mechanics applies fully.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A scientist wants to precisely locate an electron in an atom within an uncertainty of 0.1 nm. Using Heisenberg's uncertainty principle, calculate the minimum uncertainty in the electron's velocity.

Using Heisenberg's uncertainty principle: Δx × Δp ≥ h/4π, where Δx = 0.1 nm = 0.1 × 10^-9 m = 10^-10 m.
Δp ≥ h/(4π × Δx) = (6.626 × 10^-34)/(4 × 3.1416 × 10^-10) = (6.626 × 10^-34)/(1.2566 × 10^-9) = 5.27 × 10^-25 kg·m/s.
Since Δp = m × Δv: Δv ≥ Δp/m = (5.27 × 10^-25)/(9.11 × 10^-31) = 5.79 × 10^5 m/s.
An uncertainty of ~5.79 × 10^5 m/s in velocity is enormous (nearly 0.2% of the speed of light), making it impossible to define a precise trajectory for the electron; this confirms that Bohr's fixed-orbit model is fundamentally flawed for the electron.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

In Rutherford's gold foil experiment, most alpha particles passed through undeflected, but 1 in 20,000 bounced back nearly 180°. With reference to the nuclear model, explain what each observation reveals about atomic structure.

Most alpha particles passing through undeflected reveals that the atom is predominantly empty space; the electrons, with their tiny mass, cannot significantly deflect the massive, high-energy alpha particles.
A small fraction deflected at small angles indicates that within the atom there exists a region of concentrated positive charge; alpha particles passing close to this region experience repulsion and are slightly deflected.
The ~1 in 20,000 bouncing nearly 180° (head-on reflection) reveals the existence of a tiny, extremely dense, positively charged nucleus; only a direct approach to this nucleus causes such massive repulsion and back-scattering.
Calculations showed the nuclear radius is ~10^-15 m versus atomic radius ~10^-10 m — a ratio of 1:100,000; analogously, if the nucleus were a cricket ball, the atom would be about 5 km in radius, illustrating the vast empty space within atoms.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

The species X²⁻ has 18 electrons and a mass number of 32. Identify the element X, write its electronic configuration, and determine the number of protons and neutrons.

Since X²⁻ has 2 extra electrons (anion): number of protons = 18 − 2 = 16. The element with atomic number 16 is sulphur (S).
Number of neutrons = Mass number − number of protons = 32 − 16 = 16 neutrons. The species is ₁₆³²S²⁻.
Electronic configuration of S²⁻ (18 electrons): 1s² 2s² 2p⁶ 3s² 3p⁶ — isoelectronic with argon (Ar) and having a completely filled outermost shell.
This illustrates the distinction between a neutral atom's configuration and that of its ion; chemical species are characterised by atomic number (proton number), not by total electron count, and ion formation only changes the electron count.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

A photon detector in an astronomical telescope receives radiation from a star with a wavelength of 600 nm. If the total energy received is 3.15 × 10^-18 J, calculate the number of photons received.

Energy of one photon: E = hc/λ = (6.626 × 10^-34 × 3.0 × 10^8)/(600 × 10^-9) = (1.988 × 10^-25)/(6.0 × 10^-7) = 3.313 × 10^-19 J.
Number of photons received = Total energy/Energy per photon = (3.15 × 10^-18)/(3.313 × 10^-19) = 9.51 ≈ 10 photons.
This extremely small number illustrates why signals from distant stars are very weak; even a tiny amount of energy received corresponds to only a few photons when the wavelength is in the visible range.
This application demonstrates that astronomical spectroscopy relies on the particle nature of light (photon counting) and that the quantum nature of radiation is practically significant even in large-scale observations.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

The threshold frequency for caesium metal is 5.16 × 10^14 Hz. Determine whether radiation of wavelength 500 nm will cause the photoelectric effect in caesium, and if so, calculate the velocity of the ejected electron.

Frequency of incident radiation: ν = c/λ = (3.0 × 10^8)/(500 × 10^-9) = 6.0 × 10^14 Hz. Since ν (6.0 × 10^14 Hz) > ν₀ (5.16 × 10^14 Hz), photoelectric emission will occur.
Kinetic energy of ejected electron: KE = h(ν − ν₀) = 6.626 × 10^-34 × (6.0 − 5.16) × 10^14 = 6.626 × 10^-34 × 0.84 × 10^14 = 5.566 × 10^-20 J.
Using KE = ½mev²: v = √(2KE/me) = √(2 × 5.566 × 10^-20 / 9.11 × 10^-31) = √(1.222 × 10^11) = 3.50 × 10^5 m/s.
Caesium's relatively low work function makes it useful in photoelectric cells; this example shows that only the frequency of light matters for whether emission occurs, while both frequency and intensity affect the energy and number of emitted electrons respectively.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

With reference to quantum numbers, determine the possible values of n, l, and ml for electrons in the 3d subshell. How many electrons can this subshell accommodate?

For the 3d subshell: principal quantum number n = 3, azimuthal quantum number l = 2 (since d-subshell corresponds to l = 2).
Magnetic quantum number ml can take values from −l to +l: ml = −2, −1, 0, +1, +2 — giving 5 values, corresponding to 5 d-orbitals (dxy, dyz, dxz, dx²−y², dz²).
Each orbital can hold a maximum of 2 electrons with opposite spins (ms = +½ and −½); the spin quantum number ms distinguishes the two electrons within the same orbital.
Maximum electrons in 3d = 5 orbitals × 2 electrons each = 10 electrons; this can also be calculated as 2(2l+1) = 2(5) = 10; the filling of these orbitals in transition metals explains their characteristic properties.

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Chapter 2: Structure of Atom | Application Questions

A sodium lamp emitting yellow light at a wavelength of 589 nm is used in a street light. Calculate the frequency and wavenumber of this radiation and identify the region of the electromagnetic spectrum to which it belongs.

In a laboratory, students observe that when radiation of wavelength 300 nm strikes a sodium surface, electrons are emitted. The work function of sodium is 2.31 eV. Calculate whether photoelectric emission will occur, and find the kinetic energy of the emitted electrons.

A hydrogen atom in an excited state emits light when the electron falls from n = 4 to n = 2. Identify the spectral series this transition belongs to, and calculate the energy and frequency of the emitted photon.

With reference to the de Broglie hypothesis, compare the wavelengths of a golf ball (mass = 0.045 kg, velocity = 40 m/s) and an electron moving at 2 × 10^6 m/s. Comment on the significance of each wavelength.

A scientist wants to precisely locate an electron in an atom within an uncertainty of 0.1 nm. Using Heisenberg's uncertainty principle, calculate the minimum uncertainty in the electron's velocity.

In Rutherford's gold foil experiment, most alpha particles passed through undeflected, but 1 in 20,000 bounced back nearly 180°. With reference to the nuclear model, explain what each observation reveals about atomic structure.

The species X²⁻ has 18 electrons and a mass number of 32. Identify the element X, write its electronic configuration, and determine the number of protons and neutrons.

A photon detector in an astronomical telescope receives radiation from a star with a wavelength of 600 nm. If the total energy received is 3.15 × 10^-18 J, calculate the number of photons received.

The threshold frequency for caesium metal is 5.16 × 10^14 Hz. Determine whether radiation of wavelength 500 nm will cause the photoelectric effect in caesium, and if so, calculate the velocity of the ejected electron.

With reference to quantum numbers, determine the possible values of n, l, and ml for electrons in the 3d subshell. How many electrons can this subshell accommodate?

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