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Calculate the wavelength of a) a \(2.00 \mathrm{eV}\) photon, and b) an electron with kinetic energy \(2.00 \mathrm{eV}\).

Short Answer

Expert verified
The wavelength of the 2.00 eV photon is approximately 6.215 x 10^-7 m, and the wavelength of the electron with kinetic energy of 2.00 eV is approximately 3.77 x 10^-10 m.

Step by step solution

01

Conversion of energy from eV to Joules

The energy given is in electron volts (eV). We need to convert it to joules (J) to use it in the appropriate formulas. Use the conversion factor: \(1 \mathrm{eV} = 1.60218 \times 10^{-19} \mathrm{J}\). \(2.00 \mathrm{eV} \times \frac{1.60218 \times 10^{-19} \mathrm{J}}{1 \mathrm{eV}} = 3.20436 \times 10^{-19} \mathrm{J}\)
02

Wavelength of the photon

Now that we have the energy in joules, we can use the Planck-Einstein relation to calculate the wavelength of the photon: \(E = h\frac{c}{\lambda}\), where \(E\) is the energy of the photon, \(h\) is the Planck constant (\(6.62607 \times 10^{-34} \mathrm{Js}\)), \(c\) is the speed of light (\(2.998\times 10^8 \mathrm{m/s}\)), and λ is the wavelength. Rearrange the formula to solve for λ: \(\lambda = \frac{hc}{E}\) Now, substitute the values: \(\lambda = \frac{(6.62607 \times 10^{-34} \mathrm{Js})(2.998\times 10^8 \mathrm{m/s})}{3.20436 \times 10^{-19} \mathrm{J}} = 6.21541 \times 10^{-7} \mathrm{m}\) The wavelength of the \(2.00 \mathrm{eV}\) photon is approximately \(6.215\times 10^{-7} \mathrm{m}\).
03

Wavelength of the electron

We need to use the de Broglie wavelength formula to calculate the wavelength of the electron: \(\lambda = \frac{h}{p}\), where \(\lambda\) is the wavelength, \(h\) is the Planck constant, and \(p\) is the momentum of the electron. However, we don't have the momentum; we have the kinetic energy. To find the momentum, we can use the following relation between kinetic energy and momentum: \(E_K = \frac{p^2}{2m}\), where \(E_K\) is the kinetic energy, \(p\) is the momentum, and \(m\) is the mass of the electron (\(9.109\times 10^{-31} \mathrm{kg}\)). Rearrange the formula to solve for \(p\): \(p = \sqrt{2mE_K}\) Now, substitute the values: \(p = \sqrt{2(9.109\times 10^{-31} \mathrm{kg})(3.20436 \times 10^{-19} \mathrm{J})} = 1.75834 \times 10^{-24} \mathrm{kg\cdot m/s}\) Now that we have the momentum, we can use the de Broglie formula to calculate the wavelength: \(\lambda = \frac{h}{p} = \frac{6.62607 \times 10^{-34} \mathrm{Js}}{1.75834 \times 10^{-24} \mathrm{kg\cdot m/s}} = 3.76919 \times 10^{-10} \mathrm{m}\) The wavelength of an electron with kinetic energy \(2.00 \mathrm{eV}\) is approximately \(3.77 \times 10^{-10} \mathrm{m}\).

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Key Concepts

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

Planck-Einstein Relation
The Planck-Einstein relation is a fundamental principle in quantum mechanics that links the energy of a photon with its frequency. This relationship is expressed as \( E = h u \), where \( E \) is the energy, \( h \) is the Planck constant \( (6.62607 \times 10^{-34} \mathrm{Js}) \), and \( u \) is the frequency of the photon. For practical purposes, it is often more useful to relate energy to wavelength using the speed of light \( c \), resulting in the equation \( E = h\frac{c}{\lambda} \).
This equation is extremely useful because it allows us to calculate the wavelength \( \lambda \) of a photon if its energy \( E \) is known. In the process of conversion, the energy must often be converted from electron volts (eV) to joules (J) as used in international scientific units. This concept is critical in understanding how energy and light are intrinsically related and is leveraged in calculations involving photons in fields ranging from astrophysics to chemistry.
de Broglie Wavelength
The de Broglie wavelength concept extends wave-particle duality to matter, suggesting that every moving object has a wave associated with it. The de Broglie wavelength \( \lambda \) of an object is given by \( \lambda = \frac{h}{p} \), where \( h \) is the Planck constant and \( p \) is the momentum of the object.
This concept was revolutionary as it implied that not just light, but all matter, including electrons and other particles, exhibits wave-like behavior. When calculating the wavelength of an electron with a certain kinetic energy, we first need to determine its momentum. This requires an understanding of kinetic energy and its relation to momentum: \( E_K = \frac{p^2}{2m} \), where \( E_K \) is the kinetic energy and \( m \) is the mass of the particle.
By rearranging this formula to solve for momentum \( p \), and subsequently using the de Broglie equation, we can find the wavelength of the particle. This approach is essential in fields like quantum mechanics and in the operation of technologies such as electron microscopes.
Energy Conversion
Energy conversion is a routine process in scientific calculations, particularly when transitioning between different units. In physics, energy is often measured in joules (J), but in atomic and subatomic physics, electron volts (eV) are more commonly used. One electron volt is the amount of kinetic energy gained by an electron as it is accelerated through an electric potential difference of one volt.
To convert from eV to J, the conversion factor \( 1 \mathrm{eV} = 1.60218 \times 10^{-19} \mathrm{J} \) is used. Understanding this conversion is crucial when calculating quantities like wavelength from energy values.
This step ensures that energy values are in the correct form to be substituted into formulas influenced by the Planck constant or related constants ensuring consistency and correctness in calculations within physics and engineering domains.
Kinetic Energy of Electrons
Kinetic energy refers to the energy that an object possesses due to its motion, and for an electron, this energy can significantly influence its behavior at quantum levels. The kinetic energy of an electron can be mathematically expressed as \( E_K = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron and \( v \) is its velocity.
In quantum mechanics, another approach often used is connecting kinetic energy with momentum through the equation \( E_K = \frac{p^2}{2m} \). This is particularly useful when direct measurement of velocity is complex or unfeasible.
In the context of calculating wavelengths through the de Broglie hypothesis, knowing the kinetic energy is crucial. It allows us to determine the momentum of the electron, which can then be used to find its associated wavelength using the de Broglie wavelength formula. This concept forms the basis for understanding how particles behave as waves and their applications in technology and scientific research.

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Most popular questions from this chapter

Given that the work function of tungsten is \(4.55 \mathrm{eV}\) what is the stopping potential in an experiment using tungsten cathodes at \(360 \mathrm{nm} ?\)

After you told him about de Broglie's hypothesis that particles of momentum \(p\) have wave characteristics with wavelength \(\lambda=h / p\), your 60.0 -kg roommate starts thinking of his fate as a wave and asks you if he could be diffracted when passing through the 90.0 -cm-wide doorway of your dorm room. a) What is the maximum speed at which your roommate can pass through the doorway in order to be significantly diffracted? b) If it takes one step to pass through the doorstep, how long should it take your roommate to make that step (assume the length of his step is \(0.75 \mathrm{~m}\) ) in order for him to be diffracted? c) What is the answer to your roommate's question? Hint: Assume that significant diffraction occurs when the width of the diffraction aperture is less that 10.0 times the wavelength of the wave being diffracted.

How many photons per second must strike a surface of area \(10.0 \mathrm{~m}^{2}\) to produce a force of \(0.100 \mathrm{~N}\) on the surface, if the photons are monochromatic light of wavelength \(600 . \mathrm{nm}\) ? Assume the photons are absorbed.

Which of the following has the smallest de Broglie wavelength? a) an electron traveling at \(80 \%\) the speed of light b) a proton traveling at \(20 \%\) the speed of light c) a carbon nucleus traveling at \(70 \%\) the speed of light d) a helium nucleus traveling at \(80 \%\) the speed of light e) a lithium nucleus traveling at \(50 \%\) the speed of light

Consider the equivalent of Compton scattering, but the case in which a photon scatters off of a free proton. a) If \(140 .-\mathrm{keV} \mathrm{X}\) -rays bounce off of a proton at \(90.0^{\circ},\) what is their fractional change in energy \(\left(E_{0}-E\right) / E_{0} ?\) b) What energy of photon would be necessary to cause a \(1.00 \%\) change in energy at \(90.0^{\circ}\) scattering?

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