Problem 12 - College Physics

Question

When ultraviolet light with a wavelength of \(254 \mathrm{nm}\) falls upon a clean metal surface, the stopping potential necessary to terminate the emission of photoelectrons is \(0.181 \mathrm{~V}\). (a) What is the photoelectric threshold wavelength for this metal? (b) What is the work function for the metal?

Step-by-Step Solution

Answer

(a) Threshold wavelength: \(687.5\) nm. (b) Work function: \(2.89 \times 10^{-19}\) J.

Step 1: Understanding the given problem

We are dealing with photoelectric effect problems. We need to find two things: (a) The photoelectric threshold wavelength, and (b) the work function for the metal when subjected to ultraviolet light of wavelength \(254\, \mathrm{nm}\) with stopping potential \(0.181\, \mathrm{V}\). This involves understanding the energy balance in the photoelectric effect.

Step 2: Using Einstein's photoelectric equation

Einstein's equation for the photoelectric effect is \(E_k = hf - \phi\), where \(E_k\) is the kinetic energy of the emitted electrons, \(hf\) is the energy of the incident photons, and \(\phi\) is the work function. The stopping potential \(V_0\) relates to the maximum kinetic energy by \(E_k = eV_0\). Therefore: \[eV_0 = hf - \phi.\]

Step 3: Calculating energy of the incident photons

The energy of the incident photons \(hf\) can be calculated by the formula \(hf = \frac{hc}{\lambda}\). Here, \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J s}\), \(c\) is the speed of light \(3.00 \times 10^8 \text{ m/s}\), and \(\lambda\) is the wavelength. For \(\lambda = 254\, \text{nm} = 254 \times 10^{-9} \text{ m}\), the energy is given by: \[hf = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{254 \times 10^{-9}}.\]

Step 4: Solving for threshold wavelength

The threshold wavelength \(\lambda_0\) is when the energy of the incident photon is just equal to the work function \(\phi\), so there is no kinetic energy left. Thus, \(\phi = \frac{hc}{\lambda_0}.\) By rearranging, \(\lambda_0 = \frac{hc}{\phi}.\) We will calculate \(\phi\) first before finding \(\lambda_0\).

Step 5: Calculating work function \(\phi\)

Rearrange the photoelectric equation as \(\phi = hf - eV_0\). Substitute \(V_0 = 0.181\, \text{V}\) and \(hf \) from our previous calculation to find \(\phi\).

Step 6: Calculating the threshold wavelength \(\lambda_0\)

Using \(\phi\) calculated earlier, find \(\lambda_0 = \frac{hc}{\phi}.\) Substitute in the values for \(h\), \(c\), and the calculated \(\phi\) to determine \(\lambda_0\).

Step 7: Final values and comparison

Finally, confirm the calculated threshold wavelength and work function values make sense in the context of the problem and satisfy the physical constraints of the photoelectric effect.

Key Concepts

Threshold Wavelength

The threshold wavelength is a crucial concept in the study of the photoelectric effect. It is the maximum wavelength of light that can cause electrons to be emitted from a metallic surface. If the light's wavelength is longer than this threshold, no photoelectrons will be emitted at any intensity level. The significance of the threshold wavelength lies in its direct relation to the work function, which is the minimum energy needed to eject electrons from the surface. The formula that binds these concepts is \[ \phi = \frac{hc}{\lambda_0}, \]where \( \phi \) is the work function, \( h \) is Planck's constant, and \( c \) is the speed of light. Rearranging this formula gives the threshold wavelength:\[ \lambda_0 = \frac{hc}{\phi}. \]Knowing \( \lambda_0 \) helps us determine the type of electromagnetic radiation necessary to induce the photoelectric effect in a given material.

Work Function

The work function, \( \phi \), is a fundamental property of a metal surface that indicates the minimum energy required to eject an electron from that surface. When light of a certain frequency hits the surface, electrons are only emitted if the photon energy exceeds this work function. In equation form, the work function is given by:\[ \phi = hf - eV_0. \]Here, \( hf \) is the energy of the incident photons, computed from their frequency, and \( eV_0 \) refers to the energy needed to stop the electrons, reflecting their maximum kinetic energy.

• If \( hf > \phi \), electrons are emitted.
• If \( hf = \phi \), electrons are on the verge of being emitted with zero kinetic energy.
• If \( hf < \phi \), no electrons are emitted.

The work function is essential in understanding how different materials respond to various frequencies of light.

Ultraviolet Light

Ultraviolet (UV) light is electromagnetic radiation with wavelengths shorter than visible light, specifically ranging about 10 to 400 nanometers. This spectral region is high in energy compared to visible light, making it especially relevant in photoelectric applications. In the context of the photoelectric effect, UV light is often used because its shorter wavelengths equate to higher energy photons. These energetic photons have a greater chance of exceeding the work function of a material, thereby freeing electrons from the surface. Key characteristics of ultraviolet light include:

• Higher frequency than visible light, due to shorter wavelengths.
• Potentially dangerous to living organisms, capable of causing skin burns or cancer.
• Commonly used in technologies like sterilization devices and fluorescent lamps.

The energy potential of UV light makes it a valuable driver in experiments observing the photoelectric effect.

Einstein's Photoelectric Equation

Einstein's photoelectric equation is the cornerstone of understanding the photoelectric effect, describing the energy exchange between an incoming photon and an emitted electron. The equation is represented as:\[ E_k = hf - \phi, \]where \( E_k \) is the kinetic energy of emitted electrons, \( hf \) is the energy of the incident photons, and \( \phi \) is the work function of the material.

• \( hf \) is calculated by the product of Planck's constant and the frequency of the incoming light.
• The equation shows that photoelectrons are ejected only if the photon energy exceeds the work function.
• The difference \( hf - \phi \) gives the kinetic energy of the ejected electrons.

Einstein's insights were instrumental in confirming that light behaves not just as a wave, but also as a particle, laying the groundwork for quantum mechanics. Understanding this equation helps clarify why only photons with sufficient energy can dislodge electrons from a material's surface.