91Ó°ÊÓ

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

Step by step solution

01

- Derivation of intensity formula

Given the definition of intensity: \[ I = \frac{P}{A} \] Since power (\( P \)) is the energy delivered per second, we can write it using the energy of photons. The energy carried by a single photon is \( E = h f \), where \( h \) is Planck's constant, and \( f \) is the frequency. The power delivered by \( N \) photons is \( P = N h f \). Photon flux density \( \Phi \) is defined as the number of photons per second per unit area, so \( N = \Phi \times A \). Thus, \( P = \Phi h f \times A \). Substituting this in the definition of intensity: \[ I = \frac{P}{A} = \frac{ \Phi h f \times A}{A} = \Phi h f \] Therefore, \[ I = \Phi h f \]

02

- Calculating intensity for a given wavelength and photon flux density

To calculate the intensity, use the formula: \[ I = \Phi h f \] First, find the frequency \( f \) from the given wavelength \( \lambda = 5.0 \times 10^{-7} m \). Using the relation \( c = \lambda f \), where \( c \) is the speed of light, \[ f = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{5.0 \times 10^{-7}} = 6.0 \times 10^{14} \text{ s}^{-1} \] The intensity is then: \[ I = 3.8 \times 10^{18} \times 6.63 \times 10^{-34} \times 6.0 \times 10^{14} = 1.5078 \times 10^{-1} \text{ W/m}^2 \] Therefore, \[ I \approx 0.15 \text{ W/m}^2 \]

03

- Calculating the new photon flux density for a different wavelength

For the wavelength \( \lambda = 4.0 \times 10^{-7} m \), we keep the intensity the same: \[ I = 0.15 \text{ W/m}^2 \] First, find the frequency for the new wavelength: \[ f = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{4.0 \times 10^{-7}} = 7.5 \times 10^{14} \text{s}^{-1} \] Using the intensity formula \( I = \Phi h f \): \[ 0.15 = \Phi \times 6.63 \times 10^{-34} \times 7.5 \times 10^{14} \] Solving for \( \Phi \): \[ \Phi = \frac{0.15 }{ 6.63 \times 10^{-34} \times 7.5 \times 10^{14 } } = 3.0 \times 10^{18} \text{ m}^{-2} \text{s}^{-1} \] Therefore, the new photon flux density is \( \Phi = 3.0 \times 10^{18} \text{ m}^{-2} \text{s}^{-1} \)

04

- Explaining the effect on electron emission

For a light of wavelength \( 4.0 \times 10^{-7} m \) to have the same intensity as light of wavelength \( 5.0 \times 10^{-7} m \), the photon flux density must be lower. This means fewer photons hit the surface per second, resulting in fewer electrons being emitted per second.

05

- Stating the assumption

The assumption made is that each incident photon has enough energy to emit an electron and that the efficiency of electron emission remains constant for both wavelengths.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Photon Flux Density

Photon flux density, represented by the symbol \( \Phi \), is a way to quantify the number of photons that hit a surface per second per unit area. In simpler terms, it tells us how densely packed the photons are over a certain surface. Imagine shining a flashlight on a wall. The more photons that hit the wall per second in a given area, the higher the photon flux density.
The formula for photon flux density is:
\[ \Phi = \frac{N}{A} \]
where:

• \( \Phi \) is the photon flux density
• \( N \) is the number of photons
• \( A \) is the area of the surface

Photon flux density is crucial in understanding the intensity of light because the more photons that hit an area, the more energy is delivered, hence, higher intensity.

Frequency of Light

The frequency of light, denoted by \( f \), is the number of wave cycles that pass a point per second. It’s measured in Hertz (Hz). Think of it like the speed of waves in the ocean; higher frequency means more waves pass by in a shorter time.
For light, frequency is directly related to its color and energy. The formula linking frequency to the wavelength \( \lambda \) and the speed of light \( c \) is:
\[ f = \frac{c}{\lambda} \]
where:

• \( f \) is the frequency
• \( c \) is the speed of light (approximately \( 3.0 \times 10^8 \text{ m/s}\))
• \( \lambda \) is the wavelength

Understanding frequency helps us calculate the energy of photons since energy is directly proportional to frequency. This is crucial in solving problems related to the intensity of light.

Planck's Constant

Planck's constant, represented by \( h \), is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. Its value is \( 6.63 \times 10^{-34} \text{ J}\cdot\text{s} \).
The energy \( E \) of a photon can be calculated using:
\[ E = h f \]
where:

• \( E \) is the energy
• \( h \) is Planck's constant
• \( f \) is the frequency

This relationship shows that higher frequency light, like blue or ultraviolet, has more energy per photon compared to lower frequency light, like red or infrared. Planck's constant is key to linking macroscopic properties of light to its microscopic interactions with matter, making it essential in topics like the photoelectric effect and radiation intensity.