Problem 27 - University Physics with Modern Physics

Question

An ultrashort pulse has a duration of 9.00 fs and produces light at a wavelength of 556 \(\mathrm{nm} .\) What are the momentum and momentum uncertainty of a single photon in the pulse?

Step-by-Step Solution

Answer

The momentum is \(1.19 \times 10^{-27}\) kg⋅m/s; uncertainty: \(7.78 \times 10^{-29}\) kg⋅m/s.

Step 1: Convert units for pulse duration

The pulse duration is given as 9.00 femtoseconds (fs), which needs to be converted to seconds (s) because SI units are required for calculations. 1 fs = \(10^{-15}\) s, therefore 9 fs = \(9 \times 10^{-15} \) s.

Step 2: Calculate the frequency of the light

The frequency \( u \) of light can be calculated using the speed of light \( c = 3.00 \times 10^8 \) m/s and the wavelength \( \lambda = 556 \) nm = \( 556 \times 10^{-9} \) m. Use the formula:\[ u = \frac{c}{\lambda} \]Substitute the given values to find:\[ u = \frac{3.00 \times 10^8 \text{ m/s}}{556 \times 10^{-9} \text{ m}} \approx 5.395 \times 10^{14} \text{ Hz} \]

Step 3: Calculate the energy of a single photon

Using Planck's equation, calculate the energy \( E \) of a single photon with frequency \( u \):\[ E = h u \]where \( h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s} \) is Planck's constant. Substitute the calculated frequency:\[ E = 6.626 \times 10^{-34} \cdot 5.395 \times 10^{14} \approx 3.57 \times 10^{-19} \text{ J} \]

Step 4: Calculate the momentum of a single photon

The momentum \( p \) of a photon is related to its energy \( E \) and the speed of light \( c \) by the equation:\[ p = \frac{E}{c} \]Substitute the energy calculated in the previous step:\[ p = \frac{3.57 \times 10^{-19}}{3.00 \times 10^8} \approx 1.19 \times 10^{-27} \text{ kg}\cdot\text{m/s} \]

Step 5: Calculate the momentum uncertainty

The uncertainty in momentum \( \Delta p \) is related to the uncertainty in position \( \Delta x \), which is roughly the length of the pulse. \( \Delta x = c \times \text{pulse duration in seconds} \).\[ \Delta x = 3.00 \times 10^8 \text{ m/s} \times 9 \times 10^{-15} \text{ s} = 2.7 \times 10^{-6} \text{ m} \]Using Heisenberg's Uncertainty Principle:\[ \Delta x \Delta p \geq \frac{h}{4\pi} \]Substitute and solve for \( \Delta p \):\[ \Delta p \geq \frac{6.626 \times 10^{-34}}{4\pi \times 2.7 \times 10^{-6}} \approx 7.78 \times 10^{-29} \text{ kg}\cdot\text{m/s} \]

Key Concepts

Pulse Duration Conversion

In the study of ultrashort pulses, it's essential to express time in the proper units for precise calculations. Pulse duration is often given in femtoseconds (fs). However, since most calculations in physics are performed using the SI system, which requires time in seconds, a conversion is necessary.
To convert pulse duration from femtoseconds to seconds, we use the conversion factor: 1 fs = \(10^{-15}\) s.
For example, to convert 9.00 fs to seconds, we compute:

• 9.00 fs \( \times 10^{-15} \) s/fs = \(9.00 \times 10^{-15}\) s.

This conversion is crucial as it enables you to incorporate the pulse duration into various formulas and maintain consistency in units across different calculations.

Frequency Calculation

Light behaves both as a wave and as a particle, and one of its key characteristics is its frequency, \( u \). This is determined by the speed of light, \( c \), and the wavelength, \( \lambda \), via the equation:

• \( u = \frac{c}{\lambda} \)

For instance, if the wavelength of light is 556 nm, it must first be converted into meters to maintain SI unit consistency, resulting in 556 nm = \(556 \times 10^{-9}\) m.
Given the speed of light \( c = 3.00 \times 10^8 \) m/s, the frequency can be calculated as:

• \( u = \frac{3.00 \times 10^8}{556 \times 10^{-9}} \approx 5.395 \times 10^{14} \) Hz

Understanding how to calculate frequency is important for determining other properties of light, such as its energy or for analyzing its interaction with matter.

Heisenberg's Uncertainty Principle

Heisenberg's Uncertainty Principle is a fundamental theory in quantum mechanics stating that it's impossible to simultaneously know both the position and momentum of a particle with absolute certainty. In practice, this principle is usually expressed in terms of the position uncertainty \( \Delta x \) and momentum uncertainty \( \Delta p \):

• \( \Delta x \Delta p \geq \frac{h}{4\pi} \)

For an ultrashort pulse, the position uncertainty \( \Delta x \) can be approximated as the product of the speed of light and the pulse duration:
\( \Delta x = c \times \text{ pulse duration in seconds} \).
For example, a pulse duration of 9 fs results in:

• \( \Delta x = 3.00 \times 10^8 \times 9 \times 10^{-15} = 2.7 \times 10^{-6} \) meters.

Applying Heisenberg's principle allows us to compute the minimum uncertainty in momentum \( \Delta p \) for the pulse.

Photon Energy Calculation

The energy of a photon is derived using Planck's equation, which relates the photon's frequency \( u \) to its energy \( E \):

• \( E = h \cdot u \)

Here, \( h \) represents Planck's constant, approximately \(6.626 \times 10^{-34}\) J·s. By substituting the obtained frequency, the photon's energy can be determined.
For a frequency \( u \approx 5.395 \times 10^{14} \) Hz, the energy calculation becomes:

• \( E = 6.626 \times 10^{-34} \cdot 5.395 \times 10^{14} \approx 3.57 \times 10^{-19} \) Joules.

Understanding photon energy is key in areas such as quantum mechanics and spectroscopy. It helps calculate properties like momentum or even the type of interactions photons can undergo with matter.