This question will look at the intensity of radiation in a bit more detail

PHYSICS · Multi-part · CBSE · Physics · Dual Nature of Radiation and Matter · Dual Nature of Radiation and Matter

Problem

This question will look at the intensity of radiation in a bit more detail. The intensity of light, II, incident normally on an area AA is defined to be I=PAI = \frac{P}{A}, where PP is the power carried by the light. (1) Show that I=ΦhfI = \Phi hf, where Φ\Phi is the photon flux density, i.e. the number of photons incident on the surface per second per unit area, and ff is the frequency of the light. (2) Calculate the intensity of light of wavelength λ=5.0×10−7\lambda = 5.0 \times 10^{-7} m incident on a surface when the photon flux density is Φ=3.8×1018 m−2 s−1\Phi = 3.8 \times 10^{18} \text{ m}^{-2} \text{ s}^{-1} . (3) The wavelength of the light is decreased to λ=4.0×10−7\lambda = 4.0 \times 10^{-7} m. Calculate the new photon flux density so that the intensity of light incident on the surface is the same as that found in b. (4) Hence explain why light of wavelength λ=4.0×10−7\lambda = 4.0 \times 10^{-7} m and of the same intensity as that of light of wavelength λ=5.0×10−7\lambda = 5.0 \times 10^{-7} m will result in fewer electrons being emitted from the surface per second. (5) State *one* assumption made in reaching this conclusion.

Related theory

Browse the problem library