Question

In the photoelectric effect, photoelectrons begin leaving the surface at essentially the instant that light is introduced. If light behaved as a diffuse wave and an electron at the surface of a material could be assumed localized to roughly the area of an atom, it would take far longer. Estimate the time lag. assuming a work function of \(4 \mathrm{eV}\), an atomic radius of approximately \(0.1 \mathrm{nm},\) and a reasonable light intensity of \(0.01 \mathrm{~W} / \mathrm{m}^{2}\).

Short answer

The theoretical time lag for an electron to escape from the surface of the material when light behaves as a diffuse wave is approximately \(2.04 \times 10^3 s\). However, in reality, the photoelectric effect occurs almost instantaneously due to the particle behavior of light demonstrated by quantum mechanics.

Step-by-step solution

Calculate Required Energy

The energy required for the electron to escape or the work function of the material is already given in the problem as \(4 \mathrm{eV}\). One electron volt (\(eV\)) is equivalent to \(1.6 \times 10^{-19} \mathrm{J}\). Therefore, the energy required in Joules can be calculated as: \(E=4 \mathrm{eV} = 4 \times 1.6 \times 10^{-19} \mathrm{J} = 6.4 \times 10^{-19} \mathrm{J}\).

Calculate the Area of Atom's surface

The surface of the atom where the electron is localized can be modeled as a disk with radius of \(0.1 \mathrm{nm}\) or \(0.1 \times 10^{-9} \mathrm{m}\). The area of the disk can be calculated as \(A=\pi r^2 = \pi \times (0.1 \times 10^{-9} \mathrm{m})^2 = 3.14 \times 10^{-20}\ m^2\).

Calculate Theoretical Time Lag

The intensity of the light is given as \(0.01 \mathrm{W/m}^{2}\). As intensity is defined as the power delivered per unit area, the power \(P\) delivered to the area of the atom is: \(P = I \times A = 0.01 \mathrm{W/m}^{2} \times 3.14 \times 10^{-20}\ m^2 = 3.14 \times 10^{-22} \mathrm{W}\). Since \(1 \mathrm{W} = 1 \mathrm{J/s}\), this is equivalent to \(3.14 \times 10^{-22}\ \mathrm{J/s}\) energy is deposited per second. Therefore, the time required for light to deposit enough energy to free the electron is: \(t = E / P = (6.4 \times 10^{-19}\ \mathrm{J}) / (3.14 \times 10^{-22}\ \mathrm{J/s}) = 2.04 \times 10^3 s\).

Key concepts

Work Function

The work function is a crucial parameter in the study of the photoelectric effect. It represents the minimum energy required to eject an electron from the surface of a material. In simpler terms, it is like the energy barrier an electron has to overcome to leave the material. For most metals, this work function is expressed in electron volts (eV).

• In this problem, the work function is given as 4 eV, which is typical for metals used in photoelectric experiments.
• The higher the work function, the more energy is needed to free the electron.

Understanding work function helps in grasping why only certain frequencies of light can cause the photoelectric effect, since only photons with sufficient energy (higher than the work function) can eject electrons.

Atomic Radius

The atomic radius is a measure of the size of an atom, specifically, it is the distance from the nucleus to the boundary of the surrounding electron cloud. In the context of the photoelectric effect, the atomic radius helps to determine the surface area over which light is incident.

• The problem assumes an atomic radius of 0.1 nm, a typical size for atoms.
• This radius is used to calculate the area of the atom's surface that is illuminated.

By modeling the atom as a disk, we use the formula for the area of a circle, \(A = \pi r^2\), to find the surface area. This helps us understand how much of the incoming light affects an individual atom, impacting the time lag estimation for the photoelectron emission.

Light Intensity

Light intensity refers to the power per unit area received by a surface. It is an essential factor in determining how much energy is delivered to a given area from a light source.

• In the exercise, the light intensity is specified as 0.01 W/m², which is a typical value for low-intensity light.
• Light intensity directly influences the amount of energy transferred to the electron on the atom’s surface.

If the intensity were higher, more energy would be available to electrons, potentially allowing them to overcome the work function more quickly. Thus, light intensity plays a direct role in affecting the efficiency and speed of the photoelectric effect.

Energy Conversion

Energy conversion is at the heart of the photoelectric effect. It describes how light energy (photon energy) is transformed into kinetic energy of electrons. The process starts when the energy from photons, carried by light, is absorbed by electrons in a material.

• In this problem, energy conversion occurs when photons impart enough energy (at least equal to the work function) to free electrons.
• This energy conversion is quantified by measuring how much energy is needed to overcome the work function, which, in this case, is 6.4 x 10^-19 Joules.

Efficient energy conversion in the photoelectric effect leads to immediate electron emission. Conversely, as the theoretical time lag suggests, if light were a diffuse wave, energy conversion and electron emission would be significantly delayed, demonstrating the particle nature of light.