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Chapter 38: Problem 48

You hold in your hands both a green \(543-n m, 5.00-m W\) laser and a red, \(633-\mathrm{nm}, 4.00\) -mW laser. Which one will produce more photons per second, and why?

Short Answer

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Step by step solution

01

Before calculating the number of photons for each laser, we need to convert the given wavelengths (in nm) and powers (in mW) to their respective SI units: meters (m) and watts (W). We'll do this by using the following conversion factors: - 1 nm = 10^(-9) m - 1 mW = 10^(-3) W #Step 2: Calculate the energy of one photon for each laser#

Using the energy of one photon formula, we will calculate the energy of a single photon for both the green and red lasers: Energy of one green photon = (h * c) / λ (green) Energy of one red photon = (h * c) / λ (red) #Step 3: Calculate the number of photons per second for each laser#
02

Now, we will use the power of each laser and the energy of a single photon to compute the number of photons produced per second for both the green and red lasers: Number of green photons per second = (Power of green laser)/(Energy of one green photon) Number of red photons per second = (Power of red laser)/(Energy of one red photon) #Step 4: Compare the number of photons produced per second for both lasers#

Once we have calculated the number of photons per second for each laser, we will compare these numbers and determine which laser produces more photons. If the number of green photons per second > number of red photons per second, the green laser produces more photons. If the number of green photons per second < number of red photons per second, the red laser produces more photons.

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Most popular questions from this chapter

Hund's rule, a component of the Aufbauprinzip (construction principle), states that as one moves across the periodic table, with increasing atomic number, the available electron subshells are filled successively with one electron in each orbital, their spins all parallel; only when all orbitals in a subshell contain one electron are second electrons, with spins opposite to the first, placed in the orbitals. Explain why the ground state electron configurations of successive elements should follow this pattern.

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