Chapter 11: Dual Nature of Radiation and Matter Application Questions | physics Class 12 CBSE

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Chapter 11: Dual Nature of Radiation and Matter Application Questions | physics Class 12 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A solar cell made of caesium (work function 2.14 eV) is exposed to sunlight containing photons of various frequencies. With reference to Einstein's photoelectric equation, explain under what condition the cell will produce photocurrent and determine the threshold wavelength above which the cell will not respond.

The solar cell will produce photocurrent only when the frequency of incident photons exceeds the threshold frequency ν₀ = φ₀/h = 2.14 eV / 6.626 × 10⁻³⁴ J s ≈ 5.16 × 10¹⁴ Hz. Photons below this frequency have insufficient energy to eject electrons regardless of intensity.
The threshold wavelength is λ₀ = c/ν₀ = (3 × 10⁸ m/s) / (5.16 × 10¹⁴ Hz) ≈ 581 nm. Incident radiation with wavelength greater than 581 nm (lower frequency, lower photon energy) will not cause photoelectric emission in caesium.
Caesium being an alkali metal has a relatively low work function, making it sensitive to visible light. Sunlight contains substantial radiation below 581 nm (blue, violet, UV), so the caesium solar cell will respond to these components and produce photocurrent proportional to their intensity.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

In a laboratory, a student uses ultraviolet light to illuminate a zinc plate and finds photoelectric emission. When the same intensity of green light is used, no emission occurs. Using the concept of threshold frequency, explain why this happens and what fundamental property of zinc is being revealed.

Zinc has a relatively high work function, requiring photons with energy exceeding φ₀ to liberate electrons. Ultraviolet light has high frequency and short wavelength, so each UV photon carries energy hν > φ₀ for zinc, making emission possible.
Green light has a lower frequency than ultraviolet. Although its photons are present in the same number per second (same intensity), each green light photon has energy hν < φ₀ for zinc. No individual photon can supply enough energy for an electron to escape, regardless of how many photons arrive.
This reveals that zinc's threshold frequency lies between the frequencies of UV light and green light. The experiment demonstrates that photoelectric emission depends on the frequency (energy per photon) of incident radiation, not its intensity, which is a direct consequence of the quantum nature of light.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A photoelectric experiment uses two light sources of the same frequency but different intensities I₁ and I₂ (I₂ > I₁) to illuminate the same metal surface. Predict and justify what changes will be observed in (a) the photocurrent and (b) the stopping potential.

Higher intensity I₂ means more photons strike the surface per second. Since each photon still carries the same energy hν (frequency is unchanged), more photoelectrons are emitted per second. Photocurrent is directly proportional to the number of photoelectrons emitted per second, so photocurrent for I₂ is greater than for I₁.
The stopping potential V₀ = (hν − φ₀)/e depends only on the frequency ν and the work function φ₀, neither of which changes when only intensity is varied. Since no individual electron gains more energy (each still absorbs one photon of the same energy), the maximum kinetic energy is unchanged.
Therefore, the stopping potential is identical for both intensities I₁ and I₂. This result is experimentally confirmed in the V₀ versus collector potential graph (Fig. 11.3 in the chapter), where different saturation currents but the same stopping potential are observed for different intensities.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

An engineer designing a night-vision security camera wants to use the photoelectric effect to detect infrared radiation. Explain whether this is feasible based on your knowledge of threshold frequency, and suggest what type of light the camera should detect instead.

Infrared radiation has very low frequency and long wavelength, meaning each infrared photon carries very little energy E = hν. For most metals and photosensitive materials, the work function φ₀ is much larger than the energy of an infrared photon, so hν < φ₀ for infrared light.
Since the photon energy is below the work function, the threshold frequency condition is not met for infrared radiation: no photoelectric emission can occur regardless of how intense the infrared source is. The camera based purely on photoelectric effect cannot detect infrared radiation.
Instead, the camera should be designed to detect visible or ultraviolet light. Alkali metals like caesium or rubidium, with lower work functions, are sensitive even to visible light. Security cameras based on photoelectric principles should use UV-sensitive or visible-light-sensitive photomaterial rather than infrared detectors.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

In Millikan's photoelectric experiment, the slope of the V₀ versus frequency graph was found to be 4.12 × 10⁻¹⁵ V s. Explain the physical significance of this slope and calculate the value of Planck's constant using the charge of electron as 1.6 × 10⁻¹⁹ C.

The slope of the V₀ versus ν graph is h/e as derived from Einstein's equation eV₀ = hν − φ₀, rearranged as V₀ = (h/e)ν − φ₀/e. This slope is universal, independent of the metal used, and directly gives the ratio of Planck's constant to the electronic charge.
Using the given slope: h/e = 4.12 × 10⁻¹⁵ V s. Therefore, h = (4.12 × 10⁻¹⁵ V s) × (1.6 × 10⁻¹⁹ C) = 6.592 × 10⁻³⁴ J s ≈ 6.59 × 10⁻³⁴ J s, which is very close to the accepted value of Planck's constant (6.626 × 10⁻³⁴ J s).
This demonstrates how Millikan used the photoelectric effect to determine Planck's constant accurately from a simple macroscopic measurement of stopping potential versus frequency, providing strong experimental validation of Einstein's photon model.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

A laser emitting monochromatic light of frequency 6.0 × 10¹⁴ Hz has a power of 2.0 × 10⁻³ W. With reference to the photon picture of light, determine the energy of each photon and the number of photons emitted per second by the laser.

In the photon picture, each photon of frequency ν carries energy E = hν. Using h = 6.63 × 10⁻³⁴ J s and ν = 6.0 × 10¹⁴ Hz: E = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ = 3.98 × 10⁻¹⁹ J per photon.
The power of the laser P = 2.0 × 10⁻³ W is the total energy emitted per second. Since each photon carries energy E, the number of photons emitted per second N = P/E = (2.0 × 10⁻³) / (3.98 × 10⁻¹⁹) ≈ 5.0 × 10¹⁵ photons per second.
This large number of photons per second illustrates why macroscopic light beams appear continuous despite being made of discrete quanta — the photon flux is so enormous that the discrete nature is undetectable in ordinary wave optics experiments.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A student observes that in a photoelectric experiment, when she moves the light source farther from the emitter plate, the photocurrent decreases but the stopping potential remains unchanged. Explain these two observations using the photon model.

Moving the light source farther from the emitter reduces the intensity of incident radiation (intensity ∝ 1/distance²). Lower intensity means fewer photons per unit area per second strike the emitter surface. Consequently, fewer photoelectrons are emitted per second, reducing the photocurrent.
The frequency of the incident light is unchanged when the source is moved farther away. Since stopping potential V₀ = (hν − φ₀)/e depends only on frequency and work function (both unchanged), the stopping potential remains the same despite the lower intensity.
This observation directly confirms Einstein's photon model: the energy of each individual photon (hν) is a function of frequency alone, not of how far the source is. The maximum kinetic energy of each photoelectron is therefore unaffected by the source distance or intensity.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

Two metals A and B have work functions 4 eV and 2 eV respectively. Monochromatic light of photon energy 3 eV is incident on both. Using Einstein's photoelectric equation, determine for which metal photoelectric emission will occur and calculate the maximum kinetic energy of the emitted electrons.

For photoelectric emission, the photon energy must exceed the work function: hν > φ₀. For metal A (φ₀ = 4 eV): 3 eV < 4 eV, so photon energy is insufficient. No photoelectric emission occurs from metal A.
For metal B (φ₀ = 2 eV): 3 eV > 2 eV, so photon energy exceeds the work function. Photoelectric emission occurs from metal B. Using Einstein's equation: Kmax = hν − φ₀ = 3 eV − 2 eV = 1 eV.
The maximum kinetic energy of the emitted photoelectrons from metal B is 1 eV (= 1.6 × 10⁻¹⁹ J). Metal A does not emit any photoelectrons at all, regardless of the intensity of the incident 3 eV light. This illustrates how work function determines which metals respond to given radiation.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

An electron and a proton are accelerated through the same potential difference V. With reference to de Broglie's relation, compare their de Broglie wavelengths and explain which has the longer wavelength.

When a particle of charge e and mass m is accelerated through potential difference V, it gains kinetic energy K = eV = p²/2m, giving momentum p = √(2meV). The de Broglie wavelength is λ = h/p = h/√(2meV).
For the same V and same charge e: λ ∝ 1/√m. Since the proton mass (mp ≈ 1.67 × 10⁻²⁷ kg) is much greater than the electron mass (me ≈ 9.11 × 10⁻³¹ kg), the proton has much larger momentum and hence shorter wavelength.
The electron has the longer de Broglie wavelength because its smaller mass results in smaller momentum for the same kinetic energy. The ratio λe/λp = √(mp/me) ≈ √(1836) ≈ 43, meaning the electron's wavelength is about 43 times larger than the proton's for the same accelerating potential.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

A scientist claims that doubling the frequency of incident radiation on a metal surface will double the maximum kinetic energy of the emitted photoelectrons. Evaluate whether this claim is correct using Einstein's photoelectric equation.

According to Einstein's equation, Kmax = hν − φ₀. If frequency is doubled to 2ν, then K'max = h(2ν) − φ₀ = 2hν − φ₀. The new Kmax is NOT simply twice the original.
Comparing: K'max = 2hν − φ₀ = 2(hν − φ₀) + φ₀ = 2Kmax + φ₀. Since φ₀ > 0, the new maximum kinetic energy is greater than twice the original: K'max > 2Kmax. The claim is incorrect.
The claim would only be true if φ₀ = 0 (no work function), which is never the case for any real metal. The work function introduces a non-linear correction, and doubling frequency always produces more than double the kinetic energy.

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Chapter 11: Dual Nature of Radiation and Matter | Application Questions

A photoelectric experiment uses two light sources of the same frequency but different intensities I₁ and I₂ (I₂ > I₁) to illuminate the same metal surface. Predict and justify what changes will be observed in (a) the photocurrent and (b) the stopping potential.

An engineer designing a night-vision security camera wants to use the photoelectric effect to detect infrared radiation. Explain whether this is feasible based on your knowledge of threshold frequency, and suggest what type of light the camera should detect instead.

In Millikan's photoelectric experiment, the slope of the V₀ versus frequency graph was found to be 4.12 × 10⁻¹⁵ V s. Explain the physical significance of this slope and calculate the value of Planck's constant using the charge of electron as 1.6 × 10⁻¹⁹ C.

A laser emitting monochromatic light of frequency 6.0 × 10¹⁴ Hz has a power of 2.0 × 10⁻³ W. With reference to the photon picture of light, determine the energy of each photon and the number of photons emitted per second by the laser.

A student observes that in a photoelectric experiment, when she moves the light source farther from the emitter plate, the photocurrent decreases but the stopping potential remains unchanged. Explain these two observations using the photon model.

Two metals A and B have work functions 4 eV and 2 eV respectively. Monochromatic light of photon energy 3 eV is incident on both. Using Einstein's photoelectric equation, determine for which metal photoelectric emission will occur and calculate the maximum kinetic energy of the emitted electrons.

An electron and a proton are accelerated through the same potential difference V. With reference to de Broglie's relation, compare their de Broglie wavelengths and explain which has the longer wavelength.

A scientist claims that doubling the frequency of incident radiation on a metal surface will double the maximum kinetic energy of the emitted photoelectrons. Evaluate whether this claim is correct using Einstein's photoelectric equation.

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