RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Physics · 11 questions · 26 marks
Light stubbornly refuses to be just a wave or just a stream of particles — the photoelectric effect proves it comes in discrete energy packets, while de Broglie showed that even matter carries a wavelength. This chapter is the doorway into quantum physics, built almost entirely around one deceptively simple equation from Einstein and one from de Broglie.
In a photoelectric experiment, when the intensity of incident light (of frequency above threshold) is increased while keeping the frequency fixed, which of the following increases?
Answer
The photoelectric current — increasing intensity at fixed frequency delivers more photons per second, ejecting more photoelectrons per second and so raising the current. Since K_max = h(ν - ν₀) depends only on frequency, neither K_max nor the stopping potential eV₀ = K_max changes with intensity, and the work function is a fixed property of the metal.
The work function of a metal is 2.0 eV. The threshold wavelength for photoelectric emission from this metal is closest to (take hc = 1240 eV nm):
Answer
620 nm — threshold wavelength λ₀ = hc/φ₀ = 1240 eV nm/2.0 eV = 620 nm. Light of longer wavelength (lower frequency, lower photon energy) than this cannot eject photoelectrons from this metal, regardless of intensity.
An electron and a proton are accelerated through the same potential difference from rest. Which particle has the larger de Broglie wavelength?
Answer
The electron, because it has smaller mass — accelerated through the same potential V, both gain the same kinetic energy eV, so momentum p = √(2meV) depends on mass through √m. The electron, being far less massive than the proton, acquires smaller momentum for the same energy, and since λ = h/p, the smaller-momentum electron has the larger de Broglie wavelength.
A photon of energy 6.0 eV strikes a metal of work function 2.5 eV. The maximum kinetic energy of the emitted photoelectron is:
Answer
3.5 eV — by Einstein's photoelectric equation, K_max = hν - φ₀ = 6.0 - 2.5 = 3.5 eV. The photon's energy is split between overcoming the work function and giving the electron kinetic energy, with any excess appearing entirely as K_max for the most energetic (surface) electrons.
Assertion (A): Photoelectric emission from a metal surface is instantaneous, occurring within about 10⁻⁹ s of the light striking it, even at very low intensity, once the frequency is above threshold. Reason (R): In the photon picture, a single photon of sufficient energy hν ≥ φ₀ can eject an electron in a single quantum event, with no need to accumulate energy over time as the classical wave picture would require.
Answer
Both A and R are true and R is the correct explanation of A — classical wave theory predicts a time lag while a weak wave slowly deposits enough energy to free an electron, which experiments do not observe. The photon model resolves this: energy arrives in one indivisible packet hν, and if that exceeds φ₀, emission happens the instant a suitable photon is absorbed.
Light of wavelength 400 nm falls on a metal surface of work function 2.0 eV. Calculate the maximum kinetic energy of the emitted photoelectrons in electron volts (take hc = 1240 eV nm).
Answer
Photon energy, E = hc/λ = 1240/400 = 3.1 eV. By Einstein's equation, K_max = E - φ₀ = 3.1 - 2.0 = 1.1 eV. Since the photon energy exceeds the work function, photoelectrons are emitted with this maximum kinetic energy.
Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 100 V. (mass of electron m = 9.1×10⁻³¹ kg, charge e = 1.6×10⁻¹⁹ C, h = 6.63×10⁻³⁴ J s)
Answer
Kinetic energy gained, eV = ½mv², so momentum p = √(2meV). p = √(2 × 9.1×10⁻³¹ × 1.6×10⁻¹⁹ × 100) p = √(2.912×10⁻⁴⁷) p = 5.40×10⁻²⁴ kg m/s. de Broglie wavelength, λ = h/p = 6.63×10⁻³⁴/5.40×10⁻²⁴ = 1.228×10⁻¹⁰ m ≈ 0.123 nm (1.23 Å).
Sketch in words the graph of stopping potential V₀ versus frequency ν for a given photosensitive metal, and explain what its slope and its intercept on the ν-axis represent.
Answer
The graph of V₀ against ν is a straight line that does not pass through the origin: it cuts the ν-axis at the threshold frequency ν₀ (where V₀ = 0) and rises linearly for ν > ν₀. From eV₀ = hν - φ₀, rearranged as V₀ = (h/e)ν - φ₀/e, the slope of the line is h/e, a universal constant independent of the metal used — this is how Millikan experimentally verified Planck's constant. The intercept on the ν-axis is the threshold frequency ν₀ = φ₀/h, which is different for each metal and depends on its work function.
State Einstein's photoelectric equation and explain how it accounts for each of the three experimentally observed features of the photoelectric effect: (i) existence of a threshold frequency, (ii) instantaneous emission, (iii) independence of stopping potential from intensity. Also solve: light of frequency 1.0×10¹⁵ Hz falls on a metal of work function 2.0 eV; find the stopping potential (h = 6.63×10⁻³⁴ J s, e = 1.6×10⁻¹⁹ C).
Answer
Einstein's photoelectric equation: hν = φ₀ + K_max, i.e. the energy of an absorbed photon is used partly to overcome the work function φ₀ that binds the electron to the metal, and the remainder appears as the electron's maximum kinetic energy K_max. (i) Threshold frequency: emission is possible only if hν ≥ φ₀; below this, hν - φ₀ would be negative, which is meaningless as a kinetic energy, so no photoelectrons are emitted regardless of how intense (how many photons per second) the light is. Threshold frequency ν₀ = φ₀/h. (ii) Instantaneous emission: because energy transfer happens in a single photon-electron collision rather than a gradual absorption, an electron either receives a whole quantum hν ≥ φ₀ instantly and is ejected, or it isn't — there is no build-up time as classical wave theory would require. (iii) Independence from intensity: increasing intensity increases the number of photons per second, and hence the number of photoelectrons per second (current), but each individual photon still carries the same energy hν determined only by frequency; so K_max = hν - φ₀ and the stopping potential eV₀ = K_max remain unchanged. Numerical: photon energy E = hν = 6.63×10⁻³⁴ × 1.0×10¹⁵ = 6.63×10⁻¹⁹ J. Converting to eV: E = 6.63×10⁻¹⁹/1.6×10⁻¹⁹ = 4.14 eV. K_max = E - φ₀ = 4.14 - 2.0 = 2.14 eV. Since eV₀ = K_max, stopping potential V₀ = 2.14 V.
Describe the Davisson-Germer experiment and explain how it provided experimental confirmation of the de Broglie hypothesis. An electron beam accelerated through 54 V was used in the original experiment; verify that the de Broglie wavelength of these electrons is close to 1.67 Å (h = 6.63×10⁻³⁴ J s, m = 9.1×10⁻³¹ kg, e = 1.6×10⁻¹⁹ C).
Answer
Description: in the Davisson-Germer experiment, a beam of electrons accelerated through a known potential difference is directed at a nickel crystal in a vacuum chamber. A movable detector measures the intensity of electrons scattered at various angles. The experimenters found sharp maxima in the scattered electron intensity at specific angles, exactly analogous to the pattern produced when X-rays diffract off the regularly spaced atomic planes of a crystal. Confirmation of de Broglie's hypothesis: the angle at which the diffraction maximum occurred could be used, via the Bragg diffraction condition for the known crystal plane spacing, to calculate an experimental wavelength for the electron beam. This value matched closely with the wavelength predicted by de Broglie's formula λ = h/p = h/√(2meV) for the same accelerating voltage, providing direct experimental proof that a beam of particles (electrons) genuinely exhibits wave behaviour, with a wavelength set by its momentum. Verification: p = √(2meV) = √(2 × 9.1×10⁻³¹ × 1.6×10⁻¹⁹ × 54) p = √(1.572×10⁻⁴⁷) = 3.965×10⁻²⁴ kg m/s. λ = h/p = 6.63×10⁻³⁴/3.965×10⁻²⁴ = 1.672×10⁻¹⁰ m = 1.67 Å. This matches the value observed experimentally by Davisson and Germer, confirming the de Broglie relation.
Read the passage and answer the questions that follow: A photoelectric cell used in an old-style light meter has a caesium cathode with work function 1.9 eV. In a laboratory test, three different light sources are shone on it one at a time: a red laser (wavelength 650 nm), a green laser (wavelength 532 nm), and a violet laser (wavelength 405 nm), each source is used at a controllable intensity. (a) Determine which of these light sources, if any, fail to produce a photoelectric current, giving photon energies in eV (use hc = 1240 eV nm). (b) For the light source(s) that do cause emission, calculate the maximum kinetic energy of the photoelectrons for the violet laser. (c) Calculate the stopping potential needed to stop even the fastest photoelectrons produced by the violet laser. (d) If the intensity of the violet laser is doubled while its wavelength is kept the same, state what changes and what stays the same, with reasons.
Answer
(a) Photon energies: E = hc/λ. Red: E = 1240/650 = 1.908 eV. Green: E = 1240/532 = 2.331 eV. Violet: E = 1240/405 = 3.062 eV. Work function is 1.9 eV. The red laser's photon energy (1.908 eV) is only marginally above 1.9 eV, so it would just barely eject electrons with negligible kinetic energy in an idealised calculation; in practice such a marginal case is highly sensitive to the exact work function value, but taken strictly by the numbers here, none of the three sources fail, since all photon energies exceed 1.9 eV. The green and violet lasers clearly cause emission with room to spare. (b) For violet: K_max = E - φ₀ = 3.062 - 1.9 = 1.162 eV ≈ 1.16 eV. (c) Since eV₀ = K_max, the stopping potential V₀ = 1.16 V. (d) Doubling the intensity of the violet laser (at fixed wavelength, hence fixed frequency and fixed photon energy) doubles the number of photons striking the cathode per second, so the number of photoelectrons emitted per second doubles, and the photoelectric current doubles. However, K_max and the stopping potential are set only by hν - φ₀, which involves no dependence on intensity, so both remain exactly the same, at 1.16 eV and 1.16 V respectively.
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