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Chapter 39: Problem 48

An atom with mass \(m\) emits a photon of wavelength \(\lambda\). (a) What is the recoil speed of the atom? (b) What is the kinetic energy \(K\) of the recoiling atom? (c) Find the ratio \(K / E\), where \(E\) is the energy of the emitted photon. If this ratio is much less than unity, the recoil of the atom can be neglected in the emission process. Is the recoil of the atom more important for small or large atomic masses? For long or short wavelengths? (d) Calculate \(K\) (in electron volts) and \(K / E\) for a hydrogen atom (mass \(1.67 \times 10^{-27} \mathrm{~kg}\) ) that emits an ultraviolet photon of energy \(10.2 \mathrm{eV}\). Is recoil an important consideration in this emission process?

Short Answer

Expert verified
The atom's recoil speed is given by \(v=h /(m \lambda)\), and the kinetic energy is \(K= h^2 / (2m^2 \lambda^2)\), thus making \(K / E = h / (2 m c \lambda)\). The ratio \(K / E\) is much less than unity for heavier atoms and shorter wavelengths. In case of a hydrogen atom emitting an ultraviolet photon of energy 10.2 eV, the recoil does not contribute significantly to the emission process as \(K / E\) is very small.

Step by step solution

01

Find the recoil speed

The momentum of the photon is given by \(h/\lambda\) where \(h\) is the Planck's constant and \(\lambda\) is the wavelength. Momentum conservation gives the recoil speed \(v\) of the atom as \(v = h / (m \lambda)\) where \(m\) is the mass of the atom.
02

Solve for Kinetic Energy (K)

The kinetic energy \(K\) of the recoiling atom is given by \(\frac{1}{2} m v^2\). Substituting the value of \(v\), we get \(K = h^2 / (2m^2 \lambda^2)\).
03

Find the ratio \(K / E\)

The energy \(E\) of the photon is \(E=h c /\lambda\), where \(c\) is the speed of light. Hence, \(K / E =\frac{h}{2 m c \lambda}\). The ratio will be less than unity for heavier atoms and shorter wavelengths, such that the recoil effect can be neglected.
04

Numerical calculation

For a hydrogen atom with mass \(1.67 \times 10^{-27} \mathrm{~kg}\) emitting a photon of energy \(10.2 \mathrm{eV}\), we can calculate \(K\) and \(K / E\) using the analogous formulae from steps 2 and 3 using the energy in Joules (1 eV = 1.6 脳 10^-19 J). Following these operations, the recoil does not contribute significantly to the emission process as the ratio \(K / E\) is very small.

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Key Concepts

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

Photon Emission
When an atom emits a photon, it releases a packet of energy in the form of light or electromagnetic radiation. This process is crucial in understanding many phenomena in physics, especially in quantum mechanics and optics. A photon is a fundamental particle that carries energy proportional to its frequency.
- **Wavelength (\(\lambda\))**: It is the distance over which the wave's shape repeats and is inversely proportional to the photon's energy.- **Planck's Constant (\(h\))**: A fundamental constant used to calculate a photon's energy, represented as \(E = hu\), where \(u\) is the frequency.In the context of recoil speed, when a photon is emitted, it carries away momentum, causing the emitting atom to recoil in the opposite direction. This reliance on momentum reveals deeper insights into the interaction between light and matter.
Kinetic Energy
Kinetic energy (\(K\)) is the energy possessed by an object due to its motion. For the recoiling atom resulting from photon emission, this energy can be calculated using the formula \(K = \frac{1}{2} m v^2\), where \(m\) represents mass, and \(v\) is the velocity obtained from the momentum conservation principle.
The kinetic energy of the recoiling atom depends on:
  • The mass of the atom (\(m\)).
  • The velocity (\(v\)) which is influenced by the emitted photon's characteristics such as its wavelength and energy.
Given the formula derived in step 2, the kinetic energy can also be expressed via Planck's constant and the wavelength, tying it creatively into fundamental quantum properties.
Momentum Conservation
Momentum conservation is a pivotal principle in physics, stating that the total momentum of a closed system remains constant unless acted on by an external force. It becomes particularly significant in the scenario of photon emission as illustrated in the exercise.When a photon with momentum \(p = \frac{h}{\lambda}\) is emitted, the atom gains an equal and opposite momentum to maintain conservation. Here, \(h\) is Planck's constant, \(\lambda\) is the wavelength, and the recoil speed \(v = \frac{h}{m\lambda}\) pragmatically demonstrates this balance of forces.
Thus, the law of momentum conservation holds that despite the emission process causing movement, the system's overall momentum remains unchanged, perfectly encapsulating the interaction's elegance and complexity.
Energy Ratio
The energy ratio is an essential dimensionless quantity that provides insights into the significance of atom recoil during photon emission. Defined as \(\frac{K}{E}\), where \(K\) is the kinetic energy and \(E\) is the energy of the emitted photon, it indicates how much energy remains within the atom.
A smaller ratio (less than one) signifies that the energy transferred into recoil is negligible relative to the photon's energy. In other words, the photon's energy is dominant, allowing physicists to simplify models by neglecting the recoil. The ratio's magnitude depends on:
  • The mass of the atom, with smaller values for heavier atoms.
  • The photon's wavelength, where shorter wavelengths yield smaller ratios.
These dependencies illustrate that both atomic mass and photon characteristics are crucial in determining the energy dynamics within the system.

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Most popular questions from this chapter

A hydrogen atom is in a state with energy \(-1.51 \mathrm{eV}\). In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus?

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