Question

A photoelectric material having work-function \(\phi_{0}\) is illuminated with light of wavelength \(\lambda\left(\lambda<\frac{h c}{\phi_{0}}\right) .\) The fastest photoelectron has a de Broglie wavelength \(\lambda_{d} .\) A change in wavelength of the incident light by \(\Delta \lambda\) results in a change \(\Delta \lambda_{d}\) in \(\lambda_{d}\). Then the ratio \(\Delta \lambda_{d} / \Delta \lambda\) is proportional to \([\mathrm{A}] \quad \lambda_{d} / \lambda\) [B] \(\lambda_{d}^{2} / \lambda^{2}\) [C] \(\lambda_{d}^{3} / \lambda\) [D] \(\lambda_{d}^{3} / \lambda^{2}\)

Short answer

[D] \(\lambda_{d}^{3} / \lambda^{2}\)

Step-by-step solution

Calculate the initial kinetic energy of the photoelectron

Use the photoelectric effect equation to find the kinetic energy of the fastest photoelectron. The energy of the incident photon is given by the equation \(E = \frac{hc}{\lambda}\). The kinetic energy \(E_k\) of the photoelectron can be calculated using the work function \(\phi_{0}\) as follows: \[-\phi_{0} + \frac{hc}{\lambda} = E_k\]

Relate the kinetic energy to the de Broglie wavelength

The kinetic energy of the photoelectron can also be described by its de Broglie wavelength \(\lambda_d\) using the equation \[-E_k = \frac{h^2}{2m\lambda_d^2}\] where \(m\) is the mass of the electron and \(h\) is Planck's constant. Combining this with the energy from step 1 gives us \[-\phi_{0} + \frac{hc}{\lambda} = \frac{h^2}{2m\lambda_d^2}\].

Determine the change in de Broglie wavelength due to the change in incident light

Take the derivative with respect to \(\lambda\) to find the relationship between \(\Delta\lambda_d\) and \(\Delta\lambda\). The differentiation will give us \[-\frac{hc}{\lambda^2}d\lambda = -\frac{h^2}{m\lambda_d^3}d\lambda_d\]. From this equation, we can see that \(\Delta\lambda_d / \Delta\lambda\) is proportional to \(\lambda_d^3 / \lambda^2\).

Compare the options with the derived proportionality

Observe the options given and compare them with the derived proportionality relation from Step 3 to find the correct answer. The ratio \(\Delta\lambda_d / \Delta\lambda\) is directly proportional to \(\lambda_d^3 / \lambda^2\). Therefore, the correct answer is [D] \(\lambda_{d}^{3} / \lambda^{2}\).

Key concepts

de Broglie Wavelength

Understanding the de Broglie wavelength is key to diving into quantum mechanics and the behavior of particles at microscopic scales. It stems from the wave-particle duality principle, which postulates that every particle or quantum entity can be described as either a particle or a wave.

The formula to calculate a particle's de Broglie wavelength \( \lambda_d \) is given by:\[\lambda_d = \frac{h}{p}\] where \( h \) is Planck's constant and \( p \) is the momentum of the particle. In the context of the photoelectric effect, when photons (particles of light) strike a material, electrons are emitted. These electrons exhibit wave-like properties, and their de Broglie wavelength becomes noteworthy.

For electrons emitted as photoelectrons, their de Broglie wavelength gives us insight into the quantum mechanical description of their motion and position. Understanding the de Broglie wavelength helps us predict how these electrons will interact at the quantum level, which has profound implications in fields such as electronics and materials science.

Work Function

The work function, often symbolized as \( \phi_0 \) in physics, is a critical concept when discussing the photoelectric effect. It represents the minimum amount of energy required to remove an electron from the surface of a material.

The work function is a property intrinsic to the material itself and varies from one substance to another. It's typically measured in electron volts (eV) and can be thought of as an energy barrier that electrons need to overcome to be ejected from the material. When light with sufficient photon energy - that is, energy greater than or equal to the work function - hits the material, it gives the electrons enough energy to escape.

In mathematical terms, the energy \( E \) of an incident photon is related to its wavelength \( \lambda \) by the equation:\[ E = \frac{hc}{\lambda} \] If the energy of the photon is higher than the work function, the excess energy is converted into the kinetic energy of the emitted electron. This is the core principle behind the photoelectric effect and is essential for technologies like solar cells and photocathodes.

Kinetic Energy of Photoelectrons

When diving into the photoelectric effect, we encounter the kinetic energy of photoelectrons as a paramount concept. The kinetic energy of a photoelectron indicates how much energy it has as a result of its motion, which it gains after absorbing enough energy from incident photons to overcome the work function of the material.

The kinetic energy \( E_k \) of the emitted photoelectrons is given by the difference between the photon's energy and the work function:\[ E_k = \frac{hc}{\lambda} - \phi_0 \] where \( hc/\lambda \) is the energy of the photon and \( \phi_0 \) is the work function. It’s this kinetic energy that can be converted to electric current in photovoltaic cells or cause other observable effects in various applications.

Understanding how the kinetic energy of photoelectrons changes with different wavelengths of incident light is important for tuning devices like photodetectors and for experiments that reveal the quantum nature of light and matter. Variations in this kinetic energy are also closely related to changes in the de Broglie wavelength of electrons, linking back to their quantum mechanical properties.