A student documents the energy and wavelength of various electromagnetic waves in the presented table. (The term "eV" denotes electron volts, a unit of energy.) For the question mark, which electromagnetic wave should the student insert? (1 point) Responses Microwave Microwave Radio Radio Gamma Ray Gamma Ray Infrared Wave Infrared Wave Skip to navigation
To determine which electromagnetic wave should be inserted, we can use the relationship between energy and wavelength of electromagnetic waves, which is given by the equation:
E = h * c / λ
Where:
E is the energy of the wave
h is the Planck's constant (approximately 6.626 x 10^-34 J·s)
c is the speed of light (approximately 3.00 x 10^8 m/s)
λ is the wavelength of the wave
Since energy is given in electron volts (eV), we need to convert it to joules (J) using the conversion factor:
1 eV = 1.602 x 10^-19 J
Let's calculate the energy for each given wavelength:
1. Microwave (wavelength: 1 mm)
E = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (1 mm * 10^-3 m/mm) = 1.988 x 10^-24 J ≈ 12.4 meV
2. Radio (wavelength: 10 km)
E = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (10 km * 10^3 m/km) = 1.988 x 10^-25 J ≈ 124 μeV
3. Gamma Ray (wavelength: 1 nm)
E = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (1 nm * 10^-9 m/nm) = 1.988 x 10^-15 J ≈ 12.4 keV
4. Infrared Wave (wavelength: 1 μm)
E = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (1 μm * 10^-6 m/μm) = 1.988 x 10^-20 J ≈ 124 meV
Comparing the calculated energies to the given energies in the table, we find that the missing energy is closest to 12.4 meV, which corresponds to a Gamma Ray. Therefore, the student should insert "Gamma Ray" in the table.