Class 12 Physics Chapter 11: Dual Nature of Radiation and Matter case study question with answers
One case-based question of the kind the board asks: a passage and a figure, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
Show answer
Answer: (b) 1 and 2
For a given emitter the stopping potential depends only on the frequency. Curves 1 and 2 share the stopping potential −1.2 V; curve 3 stops at −2.0 V, so it was taken at a different (higher) frequency.
Show answer
Answer: (d) 2.0 eV
Kmax = eV₀ = e × 2.0 V = 2.0 eV (3.2 × 10⁻¹⁹ J). 1.2 eV belongs to curves 1 and 2; 0.8 eV is the difference of the two stopping potentials; 3.2 is the value in units of 10⁻¹⁹ J.
Answer and marking scheme
hν = Kmax + φ₀ = 2.0 + 2.3 = 4.3 eV = 4.3 × 1.6 × 10⁻¹⁹ = 6.88 × 10⁻¹⁹ J. λ = hc/E = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(6.88 × 10⁻¹⁹) ≈ 2.9 × 10⁻⁷ m = 289 nm, in the ultraviolet.
- hν = eV₀ + φ₀ = 4.3 eV (1)
- λ ≈ 289 nm (1)
Answer and marking scheme
Equal saturation currents mean equal numbers of electrons emitted per second, so curves 2 and 3 were taken at the same intensity (different frequencies). For curve 1: hν = eV₀ + φ₀ = 1.2 + 2.3 = 3.5 eV, ν = 3.5 × 1.6 × 10⁻¹⁹/6.63 × 10⁻³⁴ ≈ 8.4 × 10¹⁴ Hz.
- same intensity for 2 and 3 (1)
- ν ≈ 8.4 × 10¹⁴ Hz (1)