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You are conducting a photoelectric effect experiment by shining light of \(500 \mathrm{nm}\) wavelength a a piece of metal and determining the stopping potential. If, unbeknownst to you, your 500 nm light source actually contained a small amount of ultraviolet light, would it throw off your results by a small amount or by quite a bit? Explain.

Short Answer

Expert verified
The presence of ultraviolet light in the light source will significantly change the results, because the stopping potential is sensitive to the energy of the incident light, and ultraviolet light has higher energy due to its shorter wavelength compared to the initial 500 nm light.

Step by step solution

01

Understanding the photoelectric effect

First, it's essential to understand the photoelectric effect, especially in regards to the wavelength of the light used. According to this phenomenon, higher frequency (lower wavelength) light results in electrons with higher kinetic energy. The kinetic energy of these electrons determines the stopping potential.
02

Analysing the impact of ultraviolet light

Next, we need to consider the properties of ultraviolet light. Ultraviolet light has a shorter wavelength than the 500 nm light initially used in the experiment. Consequently, the energy of the UV light is greater, thus the kinetic energy of ejected electrons would be higher. This effect would raise the overall observed stopping potential.
03

Assesing the Effect on the Results

Having considered this, we can conclude that introducing ultraviolet light to the experiment would significantly affect the results. It won't throw off the results just by a small amount, but rather significantly, because the stopping potential, which is being measured, is sensitive to the energy of the incident light.

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Key Concepts

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

Stopping Potential
When discussing the photoelectric effect, stopping potential is a crucial concept to understand. It's the minimum electric potential needed to stop the most energetic photoelectrons from reaching the opposite electrode once they've been emitted from a material surface due to incident light.

In the absence of external light, no photoelectric emission occurs, and thus the stopping potential is zero. However, when light shines upon the material, electrons absorb energy and may be ejected. The stopping potential is then applied to counter the kinetic energy of these ejected electrons, essentially 'stopping' them. It's a direct measure of the maximum kinetic energy the electrons can have as a result of the incident light.

If ultraviolet light—which has more energy than the expected visible light—is introduced inadvertently, it can increase the kinetic energy of the emitted electrons considerably, resulting in a much higher stopping potential than anticipated. Therefore, any unintended ultraviolet light in the experiment can lead to significant errors in the measurement of stopping potential.
Wavelength of Light
The wavelength of light directly determines the energy each photon carries according to Planck's equation, which relates energy (E) to the frequency (u) of the light by the equation: \( E = hu \), where h is Planck's constant. Frequency (u) and wavelength (λ) are inversely related by the speed of light (c) using \( u = \frac{c}{\lambda} \).

So, for a given wavelength like the one mentioned in the experiment (500 nm), there's a specific amount of energy associated with the photons. If the wavelength shortens—as with the intrusion of ultraviolet light—the photons' energy increases. In the context of the photoelectric effect, higher photon energy generally means more kinetic energy for the ejected electrons, assuming the photons have enough energy to overcome the work function of the material. Therefore, if ultraviolet light sneaks into the experiment, its shorter wavelength means more energy is being imparted to the electrons, drastically altering the experimental outcome.
Kinetic Energy of Electrons
The kinetic energy of electrons in the photoelectric effect experiment can be evaluated using the photoelectric equation: \( K_{max} = E_{photon} - W \), where \( K_{max} \) is the maximum kinetic energy of the ejected electrons, \( E_{photon} \) is the energy of the incident photons, and W is the work function of the metal surface. The work function is the minimum energy required to release an electron from the surface.

For light of the prescribed wavelength of 500 nm, there's a certain expected maximum kinetic energy for the electrons. If ultraviolet light, which has more energy because of its shorter wavelength, is mixed into the light source, it exceeds this expectation. Since the electrons can gain only a specific amount of energy from the photons, any excess energy from ultraviolet light results in an increase in the electrons' kinetic energy—potentially beyond that which can be inferred from standard calculations based on visible light. This discrepancy would explain a significant increase in the measured stopping potential and a substantial deviation in the experimental results.
Ultraviolet Light Impact
Ultraviolet (UV) light holds greater energy compared to visible light due to its shorter wavelength. The impact of introducing ultraviolet light into a photoelectric effect experiment cannot be understated. It can dramatically increase the observed maximum kinetic energy of the photoelectrons due to its higher photon energy.

When UV light hits the metal surface, it can liberate more electrons than visible light and give them enough kinetic energy to be detected even after the stopping potential has been applied. This means even a small amount of UV light can lead to a larger-than-expected number of high-energy electrons, skewing the results of the experiment significantly, as they could potentially exceed the stopping potential determined for the expected wavelength of 500 nm.
Photoelectric Effect Principles
The principles of the photoelectric effect, as outlined by Albert Einstein, explain how light can be absorbed by materials to eject electrons. It fundamentally describes how the energy of photons (light particles) is transferred to electrons in a material.

Key principles include:
  • The energy of the photons must be greater than the work function of the material to eject electrons.
  • Electrons are ejected almost instantaneously once the light is shone on the material, provided the energy criteria are met.
  • The kinetic energy of ejected electrons is directly proportional to the frequency of the incident light and independent of its intensity.
  • The number of electrons ejected is proportional to the intensity of the incoming light.

These principles lay the foundation for understanding how different wavelengths and intensities of light affect the photoelectric emission of electrons from a material. Any deviation from the standard light source, say the inclusion of ultraviolet light, will significantly disturb the expected photoelectric behavior because of its larger photon energy, thereby affecting fundamental experimental determinations like stopping potential and kinetic energy calculations.

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Most popular questions from this chapter

A photon has the same momentum as an electron moving at \(10^{6} \mathrm{~m} / \mathrm{s}\). (a) Determine the photon's wavelength \(\mathrm{N}\). (b) What is the ratio of the kinetic energies of the two?

Electromagnetic "waves" strike a single slit of I \mum width. Determine the angular full width (angle from first minimum on one side of the center to first minimum on the other) in degrees of the central diffraction maximum if the waves are (a) visible light of wavelength \(500 \mathrm{nm}\) and (b) X-rays of wavelength \(0.05 \mathrm{nm}\). (c) Which more clearly demonstrates a wave nature?

We analyze the photoelectric effect using photon energy alone. Why isn't the photon momentum a consideration? (It may help to reread the discussion of momentum and energy in connection with pair production.)

A photon and an object of mass \(m\) have the same momentum \(p\). (a) Assuming that the massive object is moving slowly, so that nonrelativistic fonnulas are valid. find in terms of \(m, p\), and \(c\) the ratio of the massive ob)ect's kinetic energy to the photon's kinetic ener \(g y\), and argue that it is small. (b) Find the same ratio found in part (a), but using relativistically correct formulas for the massive object. (Nore: \(E^{2}=p^{2} c^{2}+m^{2} c^{4}\) may be helpful.) (c) Show that the low-speed limit of the ratio of part (b) agrees with part (a) and that the high-speed limit is \(1 .\) (d) Show that at very high speod, the kinetic energy of a massive object approaches \(\rho \mathrm{c}\),

In the Compton effect, we choose the electron to be at the origin and the initial photon's direction of motion to be in the \(+x\) direction. (a) We may also choose the xy-plane so that it contains the velocities of the outgoing electron and photon. Why? (b) The incoming photon's wavelength \(\lambda\) is assumed to be known. The unknowns after the collision are the outgoing photon's wavelength and direction, \(\Lambda^{\prime}\) and \(\theta,\) and the speed and direction of the electron, \(u_{e}\) and \(\phi .\) With only three equations \(-t w_{0}\) components of momentum conservation and one of energy - we can't find all four. Equation \((3-8)\) gives \(\lambda^{\prime}\) in terms of \(\theta .\) Our lack of knowledge of \(\theta\) after the collision (without an experiment) is directly related to a lack of knowledge of something before the collision. What is it? (lmagine the two objects are hard spheres.) (c) Is it reasonable to suppose that we could know this? Explain.

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