An atom with mass m emits a photon of wavelength λ. (a) What is the recoil speed of the atom? (b

An atom with mass m emits a photon of wavelength λ. (a) What...

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 eV. Is recoil an important consideration in this emission process?

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 eV. Is recoil an important consideration in this emission process?

Key Concepts

Conservation of Momentum

When an atom emits a photon, the total momentum of the system must be conserved. This means that the momentum carried away by the photon is balanced by an equal and opposite momentum imparted to the atom, resulting in a recoil. This fundamental principle is central to understanding how emission processes affect the motion of atoms.

Photon Momentum and Energy Relationship

Photons, despite having no rest mass, possess momentum that is directly related to their energy and inversely proportional to their wavelength. The relationship p = E/c or p = h/? is crucial, as it connects the energy of the emitted photon with its momentum, a connection that is vital for calculating the recoil of the atom.

Atomic Recoil and Recoil Speed

The recoil speed of an atom after emitting a photon is determined by equating the photon’s momentum with the momentum acquired by the atom (recoil momentum = m * v). This concept is important in atomic physics as it allows one to quantify the resultant motion of the atom due to the reaction force of emission.

Recoil Kinetic Energy

The kinetic energy of a recoiling atom can be computed using the classical formula K = (1/2) m v^2, where the velocity is obtained from momentum conservation. This calculation provides insight into how much energy is transferred to the atom during the emission process, which is critical in understanding energy distributions in atomic transitions.

Ratio of Recoil Kinetic Energy to Photon Energy

Evaluating the ratio K/E (kinetic energy of the recoil to the energy of the emitted photon) serves as a measure of the significance of the recoil. If this ratio is much less than unity, the recoil energy is negligible compared to the photon energy; this concept helps in determining the validity of neglecting recoil effects in various physical processes.

Dependence on Atomic Mass and Wavelength

The extent of the recoil effect depends on both the mass of the atom and the wavelength (or energy) of the photon. A smaller atomic mass results in a larger recoil speed and kinetic energy for a given photon momentum, while shorter wavelengths (or higher photon energies) lead to greater photon momentum and hence more significant recoil. This interplay is critical in the design and interpretation of experiments in atomic and quantum physics.

Energy Unit Conversion

Converting energy from one unit to another, such as from joules to electron volts, is a key skill in atomic physics. Accurate conversion is essential when comparing theoretical predictions with experimental measurements, as different contexts and subfields often use different energy units.

Transcript

0:00 Okay.

00:02 In this problem, we have a hydrogen atom with a mass of just a proton, 1 .67 times 10 negative 27 kilograms.

00:08 And this hydrogen atom emits a photon with an energy of 10 .2 ev.

00:14 We are asked to find numerous things.

00:17 Namely, we want to find the kinetic energy of this hydrogen atom, which will recoil after firing off this electron.

00:26 And the reason why it will recoil is through conservation momentum.

00:31 If it releases an electron with some momentum, it has to have an opposite momentum to maintain conservation momentum.

00:37 So it'll have a resulting kinetic energy.

00:40 And we want to get the ratio of its kinetic energy to its total energy to determine if this recoil effect is going to be important enough to consider for a photon emitted at this energy.

00:53 So for equations, we need to know the energy and momentum of a photon.

00:58 Given in terms of the blank constant and its wavelength, we'll need to know the kinetic energy and momentum for our particle with mass, which is the simple non -relativistic expression.

01:08 And i strongly believe that non -relativistic equations will be appropriate for this particular situation, because i do not think that this mass is going to have much velocity.

01:24 So let's jump into it.

01:26 So to start off, ultimately we want to use momentum.

01:32 But in order to get the momentum of a photon, we need to know its wavelength.

01:36 So we can start by taking its energy.

01:39 E equals hc over lambda.

01:43 And we can use this to find lambda, which is hc over e.

01:51 And so numerically, we'd actually need to use planks constant in its ev form, 4 .135 times 10 to the negative 15.

02:04 Electron volts seconds, multiply by the speed of light.

02:10 3 times 10 to the 8 meters per second, all divided by the stated 10 .2 ev.

02:27 And this would give us a wavelength of about 1 .22 times 10 to the negative 7 meters.

02:42 So now we can set up conservation of momentum...

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