One application of emission in remote sensing is in the detection of forest fires. The Landsat satellite has a sensor that detects active fires by examining their emission in the wavelength region of 1.55 μm to 1.75 μm.
a. What type of light is being detected by this sensor?
b. What are the energies (in kJ/mol) for photons of light at these wavelengths?
I would call this the near infrared although this is the upper end of that region.
E = hc/wavelength in meters.
E is J/photon. Multiply by 6.02E23 to convert to J/mol photons and change that to kJ/mol.
a. The sensor on the Landsat satellite detects light in the wavelength region of 1.55 μm to 1.75 μm. This range falls within the near-infrared portion of the electromagnetic spectrum. Hence, the type of light being detected by this sensor is near-infrared light.
b. To determine the energy of a photon of light at a specific wavelength, we can use the equation:
E = hc/λ
Where E represents the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of light.
First, we need to convert the wavelength from micrometers (μm) to meters. Since 1 μm is equal to 1 x 10^-6 meters, the wavelength range can be written as 1.55 x 10^-6 m to 1.75 x 10^-6 m.
Using the energy equation, we can plug in the values:
E1 = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s) / (1.55 x 10^-6 m)
E1 ≈ 4.043 x 10^-19 J
E2 = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s) / (1.75 x 10^-6 m)
E2 ≈ 3.581 x 10^-19 J
To convert the energies from joules to kilojoules per mole, we can use Avogadro's constant (N_A = 6.022 x 10^23 mol^-1):
E1 (kJ/mol) = (4.043 x 10^-19 J) / (1000 J/kJ) × (N_A mol^-1)
E1 (kJ/mol) ≈ 6.719 x 10^4 kJ/mol
E2 (kJ/mol) = (3.581 x 10^-19 J) / (1000 J/kJ) × (N_A mol^-1)
E2 (kJ/mol) ≈ 5.953 x 10^4 kJ/mol
So, the energies of photons of light at these wavelengths are approximately 6.719 x 10^4 kJ/mol and 5.953 x 10^4 kJ/mol, respectively.