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Chapter 38: Problem 52

How fast must an electron move to have a kinetic energy equal to the photon energy of sodium light at wavelength \(588 \mathrm{~nm}\) ?

Short Answer

Expert verified
The electron must move at approximately \(8.60 \times 10^5\, \text{m/s}\).

Step by step solution

01

Calculate Photon Energy

The energy of a photon is given by the formula \( E = \frac{hc}{\lambda} \), where \( h \) 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 \( 588 \times 10^{-9} \text{ m} \). Substitute the values: \[ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{588 \times 10^{-9}} \approx 3.37 \times 10^{-19} \, \text{J} \]
02

Set Kinetic Energy Equal to Photon Energy

The kinetic energy (KE) of the electron is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the electron mass \( 9.11 \times 10^{-31} \text{ kg} \) and \( v \) is the velocity of the electron. Set this equal to the photon energy calculated in Step 1: \[ \frac{1}{2}mv^2 = 3.37 \times 10^{-19} \text{ J} \]
03

Solve for Electron Velocity

Rearrange the equation to solve for velocity \( v \): \[ v = \sqrt{\frac{2 \times 3.37 \times 10^{-19}}{9.11 \times 10^{-31}}} \] Calculate the solution: \[ v \approx \sqrt{7.39 \times 10^{11}} \approx 8.60 \times 10^5 \, \text{m/s} \]

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Key Concepts

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

Photon Energy
Photon energy is a fundamental concept in the study of light and electromagnetic radiation. It describes the energy carried by a single photon, which is the basic unit of light. When light is emitted, it travels in waves, and each wave has its associated photon energy.
  • Photon energy depends on the wavelength of the light鈥攕horter wavelengths have higher energy.
  • It is calculated using the formula: \( E = \frac{hc}{\lambda} \).
  • Here, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.
These calculations are crucial in various applications, including understanding how light interacts with matter, such as electrons. It also forms the basis for quantifying energy across the electromagnetic spectrum.
Kinetic Energy
Kinetic energy is the energy of motion, an essential concept in physics. When an object moves, it possesses kinetic energy that depends on its mass and velocity. For electrons, calculating kinetic energy helps in understanding their speed and the effect of external forces on them.
  • The formula to calculate kinetic energy is \( KE = \frac{1}{2}mv^2 \).
  • \( m \) represents mass, and \( v \) is velocity.
  • Kinetic energy increases with the square of velocity, making velocity a significant factor in these calculations.
In the context of the given problem, kinetic energy is equal to the photon energy, allowing us to find the electron's velocity.
Electron Velocity
Electron velocity is a measure of how fast an electron is moving. Understanding this concept is crucial in fields such as electronics and quantum mechanics. In many applications, determining the velocity of electrons helps us understand their behavior under different conditions.
  • The velocity is typically calculated once the kinetic energy is known.
  • To find velocity in our scenario, the electron's kinetic energy is set equal to the photonic energy, allowing us to solve for \( v \) using \( v = \sqrt{\frac{2 \cdot KE}{m}} \).
  • In atomic physics, knowing electron velocities helps in predicting interactions with electromagnetic fields and other particles.
This knowledge is essential for designing devices that rely on electron movement, such as transistors and semiconductors.
Wavelength
Wavelength is a critical parameter in defining electromagnetic waves, including light. It is the distance between two consecutive crests or troughs in a wave and is typically measured in nanometers (nm) or meters.
  • The wavelength determines the type and energy of the radiation: shorter wavelengths correspond to higher energies.
  • It plays a pivotal role in calculating photon energy using \( E = \frac{hc}{\lambda} \).
  • In our problem, we examine sodium light with a wavelength of 588 nm to relate its photon energy to electron kinetic energy.
Understanding wavelength helps in various practical applications, such as designing optical devices and studying atomic and molecular transitions.
Planck's Constant
Planck's constant is a fundamental constant in quantum mechanics, symbolized as \( h \). It describes the quantized nature of energy levels in atomic systems and is a cornerstone in the study of quantum phenomena.
  • Planck's constant links energy to frequency: \( E = h \cdot f \).
  • In calculating photon energy, it is used alongside the speed of light and wavelength in the formula \( E = \frac{hc}{\lambda} \).
  • Its value is approximately \( 6.626 \times 10^{-34} \text{ J s} \).
Planck's constant has profound implications in physics, helping us understand phenomena like the photoelectric effect, quantized wave-particle duality, and the fundamental limits of small-scale energy interactions.

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Most popular questions from this chapter

If the de Broglie wavelength of a proton is \(100 \mathrm{fm}\), (a) what is the speed of the proton and (b) through what electric potential would the proton have to be accelerated to acquire this speed?

Light of wavelength \(2.40 \mathrm{pm}\) is directed onto a target containing free electrons. (a) Find the wavelength of light scattered at \(30.0^{\circ}\) from the incident direction. (b) Do the same for a scattering angle of \(120^{\circ}\).

The existence of the atomic nucleus was discovered in 1911 by Ernest Rutherford, who properly interpreted some experiments in which a beam of alpha particles was scattered from a metal foil of atoms such as gold. (a) If the alpha particles had a kinetic energy of \(7.5 \mathrm{MeV}\), what was their de Broglie wavelength? (b) Explain whether the wave nature of the incident alpha particles should have been taken into account in interpreting these experiments. The mass of an alpha particle is \(4.00 \mathrm{u}\) (unified atomic mass units), and its distance of closest approach to the nuclear center in these experiments was about \(30 \mathrm{fm}\). (The wave nature of matter was not postulated until more than a decade after these crucial experiments were first performed.)

In a photoelectric experiment using a sodium surface, you find a stopping potential of \(1.85 \mathrm{~V}\) for a wavelength of \(300 \mathrm{~nm}\) and a stopping potential of \(0.820 \mathrm{~V}\) for a wavelength of \(400 \mathrm{~nm}\). From these data find (a) a value for the Planck constant, (b) the work function \(\Phi\) for sodium, and (c) the cutoff wavelength \(\lambda_{0}\) for sodium.

What is the wavelength of (a) a photon with energy \(1.00 \mathrm{eV}\), (b) an electron with energy \(1.00 \mathrm{eV}\), (c) a photon of energy \(1.00 \mathrm{GeV}\), and \((\mathrm{d})\) an electron with energy \(1.00 \mathrm{GeV} ?\)
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