Question

What is the wavelength (in \(\mathrm{nm}\) ) of \(\gamma\) rays whose energy is \(1.50 \mathrm{MeV} ?\)

Short answer

The wavelength is approximately 0.827 nm.

Step-by-step solution

Understand Energy-Wavelength Relationship

The wavelength (\lambda) of electromagnetic radiation can be calculated from its energy using the formula \( \lambda = \frac{hc}{E} \), where \( h \) is the Planck's constant \(6.626 \times 10^{-34} \ \mathrm{Js} \), \( c \) is the speed of light \(3.00 \times 10^8 \ \mathrm{m/s} \), and \( E \) is the energy of the photon.

Step 2: Convert Energy to and Set Up the Formula

The energy given is in \( \mathrm{MeV} \), so we need to convert it to joules. 1 \mathrm{MeV} = 1.602 \times 10^{-13} \mathrm{J}.Thus, \( 1.50 \mathrm{MeV} = 1.50 \times 1.602 \times 10^{-13} \mathrm{J} = 2.403 \times 10^{-13} \mathrm{J} \).

Calculate the Wavelength

Using the energy-wavelength relationship formula:\[ \lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{2.403 \times 10^{-13}} \].Calculating this gives \( \lambda = 8.27 \times 10^{-13} \mathrm{m} \).

Convert Wavelength to Nanometers

Since 1 \( \mathrm{nm} = 1 \times 10^{-9} \ \mathrm{m} \), convert the wavelength in meters to nanometers:\( \lambda = 8.27 \times 10^{-13} \mathrm{m} \times \frac{1 \times 10^9 \mathrm{nm}}{1 \mathrm{m}} = 0.827 \mathrm{nm} \).

Key concepts

Energy-Wavelength Relationship

The relationship between energy and wavelength is essential in understanding electromagnetic radiation, such as gamma rays. This relationship is expressed through the formula \( \lambda = \frac{hc}{E} \), where \( \lambda \) represents the wavelength, \( h \) is Planck's constant, \( c \) is the speed of light, and \( E \) is the energy of the photon.

• Planck's constant (\( h \)): A fundamental constant that plays a critical role in quantum mechanics. Its value is \( 6.626 \times 10^{-34} \, \mathrm{Js} \).
• Speed of light (\( c \)): This is the constant speed at which light travels in a vacuum, approximately \( 3.00 \times 10^8 \, \mathrm{m/s} \).

These components combine to allow us to determine the wavelength of radiation when its energy is known, and vice versa. This formula is fundamental in fields like astrophysics and quantum mechanics.

Planck's Constant

Planck's constant \( h \) is vital when studying electromagnetic waves. It relates the energy of a photon to its frequency by the equation \( E = hf \).

• Historical Significance: Max Planck first introduced this constant to solve issues related to black-body radiation. It laid the groundwork for quantum theory.
• Application: In the formula \( \lambda = \frac{hc}{E} \), Planck's constant helps bridge the gap between energy and the observable phenomenon of wavelength. This makes calculating wavelengths from known energies feasible.

Understanding Planck's constant is crucial for any study involving light, radiation, or quantum mechanics. It aids in the conversion from one form of light property to another, such as wavelength to energy.

Conversion of Units

Proper unit conversion is crucial when working with electromagnetic radiation calculations. In exercises like the energy-wavelength relationship, converting units correctly is essential to obtain the correct answer.

• Converting Energy Units: The given energy was in \( \mathrm{MeV} \), requiring conversion to \( \mathrm{joules} \).
  1 \( \mathrm{MeV} = 1.602 \times 10^{-13} \mathrm{J} \).
• Converting Wavelength Units: The calculated wavelength was in meters, but expressing it in nanometers is often more practical.
  1 \( \mathrm{nm} = 1 \times 10^{-9} \, \mathrm{m} \).

Such conversions are necessary when working with the intricate scales of quantum and astronomical phenomena, ensuring that results are accurate and interpretable.

Gamma Rays Energy Calculation

Gamma rays are a form of electromagnetic radiation with very high energy. Calculating the wavelength of gamma rays involves understanding their energy and conducting precise calculations.
Gamma rays possess the shortest wavelengths and the most energy among all electromagnetic waves.
For example, given an energy of \( 1.50 \, \mathrm{MeV} \), converting this to joules first (\( 2.403 \times 10^{-13} \mathrm{J} \)) is necessary.
Using the energy-wavelength relationship formula \( \lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{2.403 \times 10^{-13}} \), the wavelength can be calculated.
Gamma rays play a significant role in several scientific fields, from medical treatments like radiation therapy to aiding in cosmic observations.