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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\(^{-27}\)kg) that emits an ultraviolet photon of energy 10.2 eV. Is recoil an important consideration in this emission process?

Short Answer

Expert verified
Recoil is negligible for hydrogen emitting 10.2 eV photons; mass and wavelength affect recoil significance.

Step by step solution

01

Recoil Speed Calculation

To calculate the recoil speed \(v\) of the atom, use the conservation of momentum. When a photon with momentum \(p = \frac{h}{\lambda}\) (where \(h\) is Planck's constant) is emitted, it equals the momentum of the recoiling atom, \(mv = \frac{h}{\lambda}\). Solving for \(v\), we get:\[ v = \frac{h}{m\lambda} \].
02

Kinetic Energy of Recoiling Atom

The kinetic energy \(K\) of the recoiling atom can be expressed as:\( K = \frac{1}{2}mv^2 \).Substitute \(v = \frac{h}{m\lambda}\) into the equation to get:\[ K = \frac{1}{2}m\left(\frac{h}{m\lambda}\right)^2 = \frac{h^2}{2m\lambda^2} \].
03

Ratio of Kinetic Energy to Photon Energy

The energy \(E\) of the emitted photon is given by the formula \( E = \frac{hc}{\lambda} \), where \(c\) is the speed of light. To find the ratio \(\frac{K}{E}\), use:\[ \frac{K}{E} = \frac{\frac{h^2}{2m\lambda^2}}{\frac{hc}{\lambda}} = \frac{h}{2mc\lambda} \].This ratio indicates how significant the atom's recoil is compared to the photon energy.
04

Effects of Atomic Mass and Wavelength

The ratio \(\frac{K}{E} = \frac{h}{2mc\lambda}\) suggests that the recoil is more significant for small atomic masses \(m\) and short wavelengths \(\lambda\). This is because the ratio increases for smaller \(m\) and \(\lambda\), indicating a larger relative importance of recoil.
05

Numerical Calculation for Hydrogen Atom

For a hydrogen atom (mass \(1.67 \times 10^{-27}\) kg) emitting a photon of energy 10.2 eV (\(\lambda = \frac{hc}{10.2 \text{ eV}}\)), calculate the kinetic energy \(K\) and the ratio \(\frac{K}{E}\). First, convert 10.2 eV to joules and then find \(\lambda\). Substitute in the expressions:1. Calculate \(v\) using \(\lambda = \frac{hc}{10.2 \times 1.6 \times 10^{-19}}\).2. Substitute \(v\) into \(K = \frac{1}{2}mv^2\).3. Calculate \(\frac{K}{E}\) and convert \(K\) to electron volts by dividing by \(1.6 \times 10^{-19}\).
06

Importance of Recoil

Compute the numerical value of \(\frac{K}{E}\). If \(\frac{K}{E} << 1\), the recoil can be neglected. For the hydrogen atom emitting at 10.2 eV, \( \frac{K}{E} \) is of the order of \(10^{-8}\), indicating recoil is not important in this case.

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Key Concepts

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

Recoil Speed
When an atom emits a photon, the atom experiences a backward momentum due to the emission. This backward movement is what we refer to as 'recoil speed.' The principle that helps us understand recoil speed is the conservation of momentum. According to this principle, the momentum before the photon emission equals the momentum after the emission. For this scenario, the momenta of the emitted photon and the recoiling atom must balance each other out.
To find the recoil speed (\( v \)), we equate the momentum of the photon, given by \( p = \frac{h}{\lambda} \), with the momentum of the recoiling atom \( mv \). Solving for the speed, we obtain:
  • \( v = \frac{h}{m\lambda} \)
This formula clearly shows that the recoil speed inversely depends on the mass of the atom and the wavelength of the emitted photon. Smaller atomic masses and shorter wavelengths lead to a larger recoil speed, as the photon carries more momentum away.
Kinetic Energy of Recoiling Atom
Once the recoil speed is determined, we can calculate the kinetic energy of the recoiling atom. Kinetic energy represents the energy of motion and is calculated using the formula:
  • \( K = \frac{1}{2}mv^2 \)
By substituting the value of \( v = \frac{h}{m\lambda} \) into the kinetic energy formula, we get:
  • \( K = \frac{h^2}{2m\lambda^2} \)
This equation shows that the kinetic energy of the recoiling atom decreases with increasing atomic mass and emission wavelength. Therefore, lighter atoms and shorter wavelengths result in higher kinetic energy after the photon emission.
Although we can calculate the kinetic energy, its magnitude and significance in practical terms heavily depend on other factors like atomic mass and the energy of the emitted photon.
Photon Energy Ratio
The ratio of the kinetic energy of the recoiling atom to the energy of the emitted photon is a crucial factor in deciding if the atomic recoil can be ignored during the emission process. This ratio expresses how much of the total energy is retained as kinetic energy in the atom, compared to what is carried away by the photon.
We calculate this ratio \( \frac{K}{E} \) using:
  • \[ \frac{K}{E} = \frac{\frac{h^2}{2m\lambda^2}}{\frac{hc}{\lambda}} = \frac{h}{2mc\lambda} \]
If this ratio is much less than one, the recoil can be considered negligible. This means most of the energy is carried away by the photon, and the atom's backward motion is insignificant.
A smaller \( m \) and a shorter \( \lambda \) yield a larger ratio, implying that recoil is more noticeable. Hence, light atoms and emissions involving short wavelengths exhibit more significant recoil effects.
Conservation of Momentum
The conservation of momentum is a fundamental concept in physics that applies to processes like photon emission and atomic recoil. It states that the total momentum of an isolated system remains constant if no external forces act upon it.
During the emission of a photon by an atom, the system involves only the atom and the photon. Here, momentum conservation ensures that the momentum lost by the photon emission equals the momentum gained by the atom, resulting in recoil.
This principle guides us in deriving key equations, such as equating photon momentum \( \frac{h}{\lambda} \) with the atomic momentum \( mv \). This derivation is foundational for calculating the recoil speed and understanding kinetic energy dynamics.
Conservation laws facilitate not only these calculations but also help in predicting the outcomes of emissions in different atomic and irradiation conditions, ensuring consistency and predictability in physical phenomena.

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Most popular questions from this chapter

For your work in a mass spectrometry lab, you are investigating the absorption spectrum of one-electron ions. To maintain the atoms in an ionized state, you hold them at low density in an ion trap, a device that uses a configuration of electric fields to confine ions. The majority of the ions are in their ground state, so that is the initial state for the absorption transitions that you observe. (a) If the longest wavelength that you observe in the absorption spectrum is 13.56 nm, what is the atomic number Z for the ions? (b) What is the next shorter wavelength that the ions will absorb? (c) When one of the ions absorbs a photon of wavelength 6.78 nm, a free electron is produced. What is the kinetic energy (in electron volts) of the electron?

(a) In an electron microscope, what accelerating voltage is needed to produce electrons with wavelength 0.0600 nm? (b) If protons are used instead of electrons, what accelerating voltage is needed to produce protons with wavelength 0.0600 nm? (\(Hint\): In each case the initial kinetic energy is negligible.)

A 10.0-g marble is gently placed on a horizontal tabletop that is 1.75 m wide. (a) What is the maximum uncertainty in the horizontal position of the marble? (b) According to the Heisenberg uncertainty principle, what is the minimum uncertainty in the horizontal velocity of the marble? (c) In light of your answer to part (b), what is the longest time the marble could remain on the table? Compare this time to the age of the universe, which is approximately 14 billion years. (\(Hint\): Can you know that the horizontal velocity of the marble is \(exactly\) zero?)

(a) What is the energy of a photon that has wavelength 0.10 \(\mu\)m ? (b) Through approximately what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinhole 0.10 \(\mu\)m in diameter? What is the speed of these electrons? (c) If protons rather than electrons were used, through what potential difference would protons have to be accelerated so they would exhibit wave nature in passing through this pinhole? What would be the speed of these protons?

(a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the \(n\) = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels. (c) The average lifetime of the first excited level of a hydrogen atom is 1.0 \(\times\) 10\(^{-8}\) s. In the Bohr model, how many orbits does an electron in the \(n\) = 2 level complete before returning to the ground level?

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