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The wavelength of the yellow spectral emission line of sodium is \(590 \mathrm{nm}\). At what kinetic energy would an electron have that wavelength as its de Broglie wavelength?

Short Answer

Expert verified
The electron's kinetic energy is approximately \(3.318 \times 10^{-19}\) Joules.

Step by step solution

01

Understand the de Broglie wavelength formula

The de Broglie wavelength \( \lambda \) of a particle, such as an electron, is given by the equation \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \) m² kg/s, and \( p \) is the momentum of the particle.
02

Express momentum in terms of kinetic energy

The momentum \( p \) of a particle is related to its kinetic energy \( KE \) by the formula \( p = \sqrt{2m \cdot KE} \), where \( m \) is the mass of the electron \( 9.109 \times 10^{-31} \) kg.
03

Substitute the momentum expression into the de Broglie formula

Substitute \( p = \sqrt{2m \cdot KE} \) into the de Broglie formula: \( \lambda = \frac{h}{\sqrt{2m \cdot KE}} \). Rearranging for \( KE \), we get \( KE = \frac{h^2}{2m \cdot \lambda^2} \).
04

Plug in the given values

Plug in the values: \( h = 6.626 \times 10^{-34} \) m² kg/s, \( m = 9.109 \times 10^{-31} \) kg, and \( \lambda = 590 \times 10^{-9} \) m. This yields \( KE = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.109 \times 10^{-31} \times (590 \times 10^{-9})^2} \).
05

Calculate the kinetic energy

Perform the calculations: \( KE = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.109 \times 10^{-31} \times (590 \times 10^{-9})^2} \approx 3.318 \times 10^{-19} \) Joules.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a form of energy that an object possesses due to its motion. It is defined as the work required to accelerate an object of a given mass from rest to its current velocity. In mathematical terms, the kinetic energy (KE) of an object with mass (m) and velocity (v) is given by:\[ KE = \frac{1}{2} mv^2 \]For tiny particles like electrons, moving at high velocities, this energy becomes very significant. When considering the relationship between kinetic energy and momentum, the equation for momentum \( p \), which is \( p = \sqrt{2m \cdot KE} \), helps us understand how mass and velocity interplay to create kinetic energy.In the context of de Broglie wavelength, kinetic energy helps to determine the wavelength of particles like electrons. A higher kinetic energy usually corresponds to a shorter wavelength, due to the particle moving more quickly. Hence, understanding kinetic energy is crucial in fields like quantum mechanics, where the behavior of subatomic particles is studied.
Planck's Constant
Planck's constant plays a pivotal role in the foundation of quantum mechanics. Denoted as \( h \), it has the value of \( 6.626 \times 10^{-34} \) m² kg/s. This constant is fundamental in quantifying the energy of photons, as seen in the Planck-Einstein relation:\[ E = h u \]Here, \( E \) is the energy of the photon, and \( u \) (nu) is its frequency. Planck’s constant is not only crucial for photons but also appears in the calculation of the de Broglie wavelength, where it relates to the momentum of particles.The formula for de Broglie wavelength \( \lambda = \frac{h}{p} \) shows how Planck's constant bridges the gap between wave and particle physics by associating particle momentum with wavelike properties. This constant is an essential parameter for converting classical physics concepts into the quantum realm, helping to define how energy quantization occurs.
Momentum
Momentum is a measure of the motion of an object and is the product of its mass and velocity. In physics, it is denoted by \( p \) and expressed as:\[ p = mv \]This physical quantity describes how much motion an object has and its capability to continue moving unless acted on by a force.In quantum mechanics, momentum plays a fundamental role in determining the wavelength of subatomic particles through the de Broglie equation \( \lambda = \frac{h}{p} \). By expressing momentum in terms of kinetic energy using the relation \( p = \sqrt{2m \cdot KE} \), we can bridge the understanding of classical objects to quantum particles.Momentum helps in providing insights into the dynamics of microscopic systems. It's essential to calculate the de Broglie wavelength, a crucial concept when predicting the behavior of electrons, photons, and other subatomic particles in experimental and theoretical physics.
Spectral Emission Line
A spectral emission line is a unique frequency emitted by an element, observed as a bright line in a spectrometer. Each element has a distinct set of spectral lines, which arise from the transitions of electrons between different energy levels within an atom. These lines provide valuable information about the composition and properties of substances in both terrestrial and astronomical sources. They are used in spectroscopy to identify elements by their optical "fingerprint." The color of the spectral emission line, like the yellow line of sodium at 590 nm, corresponds to a specific wavelength, which can be associated with the motion and energy of particles like electrons. By employing the de Broglie wavelength concept, one can calculate related kinetic energy of electrons having specific wavelengths corresponding to spectral lines. Thus, understanding spectral emission lines not only aids in identifying elements but also plays a critical role in studying the quantum behavior of particles and analyzing their energy states within atoms.

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Most popular questions from this chapter

A \(3.0 \mathrm{MeV}\) proton is incident on a potential energy barrier of thickness \(10 \mathrm{fm}\) and height \(10 \mathrm{MeV}\). What are (a) the transmission coefficient \(T,\) (b) the kinetic energy \(K,\) the proton will have on the other side of the barrier if it tunnels through the barrier, and (c) the kinetic energy \(K\), it will have if it reflects from the barrier? A \(3.0 \mathrm{MeV}\) deuteron (the same charge but twice the mass as a proton) is incident on the same barrier. What are (d) \(T,(\mathrm{e}) K_{\mathrm{N}}\) and (f) \(K_{r} ?\)

A photon undergoes Compton scattering off a stationary free clectron. The photon scatters at \(90.0^{\circ}\) from its initial direction; its initial wavelength is \(3.00 \times 10^{-12} \mathrm{~m}\). What is the electron's kinetic energy?

The work function of tungsten is \(4.50 \mathrm{eV}\). Calculate the speed of the fastest electrons ejected from a tungsten surface when light whose photon energy is \(5.80 \mathrm{eV}\) shines on the surface.

A nonrelativistic particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is \(1.813 \times 10^{-4} .\) By calculating its mass, identify the particle.

Suppose the fractional efficiency of a cesium surface (with work function \(1.80 \mathrm{eV}\) ) is \(1.0 \times 10^{-16}\); that is, on average one electron is ejected for every \(10^{16}\) photons that reach the surface. What would be the current of electrons ejected from such a surface if it were illuminated with \(600 \mathrm{nm}\) light from a \(2.00 \mathrm{~mW}\) laser and all the ejected electrons took part in the charge flow?

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