Problem 2 - College Physics

Question

\(\cdot\) An excited nucleus emits a gamma-ray photon with an energy of 2.45 MeV. (a) What is the photon's frequency? (b) What is the photon's wavelength? (c) How does this wave-length compare with a typical nuclear diameter of \(10^{-14} \mathrm{m} ?\)

Step-by-Step Solution

Answer

The photon's frequency is approximately 5.926 × 10²⁰ Hz, wavelength is 5.06 × 10⁻¹³ m, and it's about 50.6 times larger than a typical nuclear diameter.

Step 1: Convert Energy to Joules

The energy given is in MeV, so first, we convert it into joules. 1 MeV is equal to \(1.60219 \times 10^{-13} \text{ J}\). Therefore, the energy \(E\) of the photon in joules is calculated as follows:\[E = 2.45 \times 1.60219 \times 10^{-13} \text{ J} = 3.92635 \times 10^{-13} \text{ J}\].

Step 2: Calculate Photon's Frequency

Using the formula \(E = h \cdot f\) where \(E\) is energy, \(h\) is Planck's constant \(h = 6.626 \times 10^{-34} \text{ J s}\), and \(f\) is frequency, solve for \(f\):\[f = \frac{E}{h} = \frac{3.92635 \times 10^{-13}}{6.626 \times 10^{-34}} \approx 5.926 \times 10^{20} \text{ Hz}\].

Step 3: Calculate Photon's Wavelength

Using the relationship \(c = \lambda \cdot f\), where \(c = 3.00 \times 10^8 \text{ m/s}\) is the speed of light, \(\lambda\) is wavelength, and \(f\) is frequency, solve for \(\lambda\):\[\lambda = \frac{c}{f} = \frac{3.00 \times 10^8}{5.926 \times 10^{20}} \approx 5.06 \times 10^{-13} \text{ m}\].

Step 4: Compare with Typical Nuclear Diameter

The wavelength calculated before is \(5.06 \times 10^{-13} \text{ m}\), which we compare with a typical nuclear diameter of \(10^{-14} \text{ m}\). We see: \[\frac{5.06 \times 10^{-13}}{10^{-14}} = 50.6\]Thus, the wavelength is about 50.6 times larger than a typical nuclear diameter.

Key Concepts

Photon Energy

Understanding photon energy is the first step in analyzing gamma-ray photons. In physics, energy is commonly expressed in units of electronvolts, and for higher-energy scenarios like gamma rays, we use mega electronvolts (MeV). 1 MeV is equivalent to \(1.60219 \times 10^{-13} \text{ J}\).
This relationship allows us to convert photon energy from MeV to joules to be able to perform further calculations utilizing the standard SI unit of energy.
Converting the given 2.45 MeV energy results in:

• \(E = 2.45 \times 1.60219 \times 10^{-13} \text{ J} = 3.92635 \times 10^{-13} \text{ J}\).

This conversion might seem trivial, but it's an essential step toward using the energy value in subsequent calculations like finding frequency and wavelength, due to the way scientific constants are defined.

Frequency Calculation

Calculating the frequency of a photon provides insight into its energy distribution. The formula for finding photon frequency derives from Planck's equation \(E = h \cdot f\), where \(E\) is energy, \(h\) is Planck's constant \((6.626 \times 10^{-34} \text{ J s})\), and \(f\) is the frequency.
Given the energy calculated, we plug the values into the formula:

• \(f = \frac{E}{h} = \frac{3.92635 \times 10^{-13}}{6.626 \times 10^{-34}} \approx 5.926 \times 10^{20} \text{ Hz}\).

This high frequency is indicative of the gamma-ray photons' extremely short wavelengths and high energy, distinguishing them from other electromagnetic spectrum regions such as visible light or radio waves.

Wavelength Comparison

The wavelength of a gamma-ray photon offers a peek into the scale and impact of these energetic particles. Wavelength \(\lambda\) is calculated through the equation \(c = \lambda \cdot f\), where \(c\) is the speed of light \((3.00 \times 10^8 \text{ m/s})\). Substituting the previously calculated frequency:

• \(\lambda = \frac{c}{f} = \frac{3.00 \times 10^8}{5.926 \times 10^{20}} \approx 5.06 \times 10^{-13} \text{ m}\).

The short wavelength reflects the high frequency and energy level, emphasizing the compact and powerful nature of gamma rays compared to other electromagnetic waves.

Nuclear Diameter

Gamma-ray photons have wavelengths that can be compared with the scale of atomic nuclei to understand their penetrative capability. The calculated wavelength of \(5.06 \times 10^{-13} \text{ m}\) is compared with a typical nuclear diameter of \(10^{-14} \text{ m}\). Through a simple division, the comparison becomes apparent:

• \(\frac{5.06 \times 10^{-13}}{10^{-14}} = 50.6\).

This suggests that the wavelength is approximately 50.6 times larger than the nuclear diameter.
This relative size difference implies that gamma rays have the capability to penetrate deeply into materials, making them invaluable in applications such as medical imaging and treatments, as well as in nuclear physics research.