Question

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

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

Step-by-step solution

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:\[ v = \frac{h}{m\lambda} \].

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:\[ K = \frac{1}{2}m\left(\frac{h}{m\lambda}\right)^2 = \frac{h^2}{2m\lambda^2} \].

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.

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.

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 \(1.6 \times 10^{-19}\).

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.

Key concepts

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.