Problem 11
Question
The vapours of \(\mathrm{Hg}\) absorb some electron accelerated by a potential difference of \(5.0 \mathrm{~V}\) as a result of which light is emitted. If the full energy of single incident electron is supposed to be converted into light emitted by single Hg-atom, the wavelength of the emitted light is (a) \(2480 \mathrm{~nm}\) (b) \(248 \mathrm{~nm}\) (c) \(6200 \mathrm{~nm}\) (d) \(620 \mathrm{~nm}\)
Step-by-Step Solution
Verified Answer
The correct answer is (b) 248 nm.
1Step 1: Define the Known Variables
The kinetic energy acquired by an electron due to the potential difference of 5.0 V is known, and the energy of a single electron is given by the equation \(E = eV\), where \(e\) is the elementary charge and \(V\) is the potential difference.
2Step 2: Calculate the Energy of the Electron
Use the formula \(E = eV\) to calculate the energy (in electron volts) of an electron after being accelerated through a 5.0 V potential difference. Note that the elementary charge \(e\) is approximately \(1.602 \times 10^{-19} \mathrm{C}\).
3Step 3: Calculate the Wavelength of the Emitted Light
The energy of the emitted light can be related to its wavelength by using the Planck's equation \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant and \(c\) is the speed of light. Rearrange this formula to solve for the wavelength \(\lambda\): \(\lambda = \frac{hc}{E}\).
4Step 4: Substitute the Values and Solve for Wavelength
Plug in the values for Planck's constant \(h = 6.626 \times 10^{-34} \mathrm{ J \cdot s}\), speed of light \(c = 3.00 \times 10^{8} \mathrm{ m/s}\), and energy \(E\) calculated in Step 2 into the equation to find the wavelength of the emitted light.
5Step 5: Compare the Wavelength with the Given Options
The calculated wavelength should be compared with the given options to identify the correct answer.
Key Concepts
Photoelectric EffectPlanck's EquationWavelength CalculationKinetic Energy of Electrons
Photoelectric Effect
The photoelectric effect is a phenomenon in physics where electrons are emitted from a substance (such as a metal or semiconductor) after it absorbs electromagnetic radiation, typically in the form of light. It was first observed by Heinrich Hertz in 1887, and it's a crucial concept in understanding the interaction between light and matter.
When light with sufficient energy shines upon a material, it can transfer energy to electrons within the material. If the energy of the photons (light particles) is greater than the work function (the minimum energy needed to remove an electron from the surface of the material), electrons will be ejected. This results in an observable current if the material is part of a circuit.
The photoelectric effect provided essential evidence for the quantum theory of light, suggesting that light can behave both as a wave and as a particle. Albert Einstein famously explained this phenomenon in 1905, for which he received the Nobel Prize in Physics in 1921.
When light with sufficient energy shines upon a material, it can transfer energy to electrons within the material. If the energy of the photons (light particles) is greater than the work function (the minimum energy needed to remove an electron from the surface of the material), electrons will be ejected. This results in an observable current if the material is part of a circuit.
The photoelectric effect provided essential evidence for the quantum theory of light, suggesting that light can behave both as a wave and as a particle. Albert Einstein famously explained this phenomenon in 1905, for which he received the Nobel Prize in Physics in 1921.
Planck's Equation
Planck's equation, devised by Max Planck, reveals the relationship between the energy of a photon and its frequency. The equation is represented as: \(E = hu\), where \(E\) is the energy of the photon, \(h\) is Planck's constant (\(6.626 \times 10^{-34} \mathrm{J \cdot s}\)), and \(u\) is the frequency of the photon.
This equation is foundational to quantum mechanics and shows that the energy of a photon is quantized, meaning it comes in discrete packets rather than a continuous range. One can also express the equation in terms of wavelength (\(\lambda\)) since frequency and wavelength are inversely related by the speed of light \(c\): \(E = \frac{hc}{\lambda}\).
Understanding Planck's equation is crucial for solving numerous problems in quantum physics, like calculating the energy of a photon needed to induce transitions within atoms or determining the energy of emitted or absorbed light, as illustrated in photoelectric effect problems.
This equation is foundational to quantum mechanics and shows that the energy of a photon is quantized, meaning it comes in discrete packets rather than a continuous range. One can also express the equation in terms of wavelength (\(\lambda\)) since frequency and wavelength are inversely related by the speed of light \(c\): \(E = \frac{hc}{\lambda}\).
Understanding Planck's equation is crucial for solving numerous problems in quantum physics, like calculating the energy of a photon needed to induce transitions within atoms or determining the energy of emitted or absorbed light, as illustrated in photoelectric effect problems.
Wavelength Calculation
Wavelength calculation is a fundamental concept in wave mechanics and photonics. The wavelength (\(\lambda\)) of a light wave is the distance over which the wave's shape repeats. It is inversely proportional to the frequency (\(u\)): \(\lambda = \frac{c}{u}\), where \(c\) is the speed of light.
To calculate the wavelength of light emitted when an electron transitions between energy levels in an atom, like in the photoelectric effect, Planck's equation is used in the form of \(\lambda = \frac{hc}{E}\). Here, \(E\) is the energy difference between the two levels, which is equal to the energy of the photon emitted during the transition.
Measuring the wavelength of light allows us to understand more about the energy transitions within different materials and is a vital part of spectroscopy, which is used across sciences from physics to chemistry and even astronomy.
To calculate the wavelength of light emitted when an electron transitions between energy levels in an atom, like in the photoelectric effect, Planck's equation is used in the form of \(\lambda = \frac{hc}{E}\). Here, \(E\) is the energy difference between the two levels, which is equal to the energy of the photon emitted during the transition.
Measuring the wavelength of light allows us to understand more about the energy transitions within different materials and is a vital part of spectroscopy, which is used across sciences from physics to chemistry and even astronomy.
Kinetic Energy of Electrons
The kinetic energy of electrons is the energy that an electron possesses due to its motion. In the context of the photoelectric effect problems, when electrons are emitted from a material's surface after absorbing energy from incident photons, they will have a certain kinetic energy. The maximum kinetic energy that the ejected electrons can acquire is given by the equation \(K_{\max} = hu - \Phi\), where \(hu\) is the energy of the incident photon and \(\Phi\) is the work function of the material.
The kinetic energy of the electrons also can be determined by the potential difference they traverse; this is based on the equation \(E = eV\), where \(e\) is the elementary charge and \(V\) is the potential difference. For an electron moving due to a potential difference, its kinetic energy is equal to the electric potential energy it gains while moving through that potential difference.
This understanding is not only important for investigations into the photoelectric effect but also for applications like electron microscopy and designing electronic devices where controlling the kinetic energy of electrons is essential.
The kinetic energy of the electrons also can be determined by the potential difference they traverse; this is based on the equation \(E = eV\), where \(e\) is the elementary charge and \(V\) is the potential difference. For an electron moving due to a potential difference, its kinetic energy is equal to the electric potential energy it gains while moving through that potential difference.
This understanding is not only important for investigations into the photoelectric effect but also for applications like electron microscopy and designing electronic devices where controlling the kinetic energy of electrons is essential.
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