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Chapter 6: Problem 31

A diode laser emits at a wavelength of \(987 \mathrm{~nm}\). (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 \(0.52 \mathrm{~J}\) over a period of \(32 \mathrm{~s}\). How many photons per second are being emitted by the laser?

Short Answer

Expert verified
(a) The radiation belongs to the near-infrared region of the electromagnetic spectrum. (b) The diode laser emits approximately \(8.04 \times 10^{16}\) photons per second.

Step by step solution

01

Identify the portion of the electromagnetic spectrum

The laser emits at a wavelength of \(987 nm\), which is equivalent to \(987 \times 10^{-9} m\). By referring to the electromagnetic spectrum, we can find that this wavelength belongs to the near-infrared region.
02

Calculate the energy of a single photon

First, we need to find the frequency of the emitted radiation using the speed of light equation: \(c = \nu \lambda \Rightarrow \nu = \frac{c}{\lambda}\) Plugging in the values, we get: \(\nu = \frac{3 \times 10^8 m/s}{987 \times 10^{-9} m} \approx 3.04 \times 10^{14} Hz\) Now, we can calculate the energy of a single photon using the energy of a photon equation: \(E = h\nu\) Plugging in the values, we get: \(E = (6.63 \times 10^{-34} Js)(3.04 \times 10^{14} Hz) \approx 2.02 \times 10^{-19} J\)
03

Calculate the number of photons emitted per second

To find the number of photons emitted per second, we need to use the given total energy and period: Total energy absorbed = \(0.52 J\) Period = \(32 s\) First, we find the energy absorbed per second (also known as the power): Power = \(\frac{Total~Energy}{Period}\) Power = \( \frac{0.52 J}{32 s} \approx 0.01625 J/s \) Now, we divide the energy absorbed per second by the energy of a single photon to find the number of photons emitted per second: Number of photons per second = \(\frac{Power}{Energy~per~photon}\) Number of photons per second = \( \frac{0.01625 J/s}{2.02 \times 10^{-19} J} \approx 8.04 \times 10^{16}~photon/s\) Therefore, the diode laser is emitting approximately \(8.04 \times 10^{16}\) photons per second.

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Most popular questions from this chapter

(a) A green laser pointer emits light with a wavelength of \(532 \mathrm{~nm}\). What is the frequency of this light? (b) What is the energy of one of these photons? (c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of 532 -nm photons. What is the energy gap between the ground state and excited state in the laser material?

The following electron configurations represent excited states. Identify the element and write its ground-state condensed electron configuration. (a) \(1 s^{2} 2 s^{2} 2 p^{4} 3 s^{1}\), (b) \([\) Ar \(] 4 s^{1} 3 d^{10} 4 p^{2} 5 p^{1}\), (c) \([\mathrm{Kr}] 5 s^{2} 4 d^{2} 5 p^{1}\)

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