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Ch.6 - Electronic Structure of Atoms
Chapter 6, Problem 31

A diode laser emits at a wavelength of 987 nm. (a) In what portion of the electromagnetic spectrum is this radiation found? (b) All of its output energy is absorbed in a detector that measures a total energy of 0.52 J over a period of 32 s. How many photons per second are being emitted by the laser?

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Step 1: Identify the portion of the electromagnetic spectrum where the radiation is found. The wavelength of the radiation is given as 987 nm. This falls within the near-infrared (NIR) region of the electromagnetic spectrum, which ranges from about 800 nm to 2500 nm.
Step 2: Calculate the energy of a single photon. Use the formula E = hc/λ, where 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 in meters. Remember to convert the wavelength from nm to m by multiplying by 10^-9.
Step 3: Calculate the total number of photons absorbed by the detector. Divide the total energy absorbed by the energy of a single photon. This will give you the total number of photons absorbed.
Step 4: Calculate the number of photons emitted per second. Divide the total number of photons absorbed by the total time in seconds. This will give you the number of photons emitted per second by the laser.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, arranged by wavelength or frequency. It includes gamma rays, X-rays, ultraviolet light, visible light, infrared radiation, microwaves, and radio waves. A wavelength of 987 nm falls within the infrared region, which is just beyond the visible spectrum, indicating that this radiation is not visible to the human eye but can be detected by specialized instruments.
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Photon Energy

Photons are particles of light that carry energy, which can be calculated using the equation E = hc/λ, where E is energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. For the diode laser emitting at 987 nm, this equation allows us to determine the energy of a single photon. Understanding photon energy is crucial for calculating the number of photons emitted based on the total energy detected.
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Photon Emission Rate

The photon emission rate refers to the number of photons emitted per unit time, typically expressed in photons per second. To find this rate, one can divide the total energy detected by the energy of a single photon and then divide by the time period over which the energy was measured. This concept is essential for quantifying the output of the laser in terms of its photon production, which is relevant in various applications, including telecommunications and medical devices.
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Related Practice
Textbook Question

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325 nm. (c) How many photons are in a 1.00 mJ burst of this radiation?

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Textbook Question

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325 nm. (d) These UV photons can break chemical bonds in your skin to cause sunburn—a form of radiation damage. If the 325-nm radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in kJ>mol.

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Textbook Question

The energy from radiation can be used to cause the rupture of chemical bonds. A minimum energy of 242 kJ/mol is required to break the chlorine–chlorine bond in Cl2. What is the longest wavelength of radiation that possesses the necessary energy to break the bond? What type of electromagnetic radiation is this?

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Textbook Question

A stellar object is emitting radiation at 3.55 mm. a. What type of electromagnetic spectrum is this radiation? b. If a detector is capturing 3.2×108 photons per second at this wavelength, what is the total energy of the photons detected in 1.0 hour?

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Textbook Question

Molybdenum metal must absorb radiation with a minimum frequency of 1.09 * 1015 s - 1 before it can eject an electron from its surface via the photoelectric effect. (a) What is the minimum energy needed to eject an electron?

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Textbook Question

Molybdenum metal must absorb radiation with a minimum frequency of 1.09 * 1015 s - 1 before it can eject an electron from its surface via the photoelectric effect. (b) What wavelength of radiation will provide a photon of this energy?

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