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Chapter 1: Problem 40

If a cluster can be broken up by a photon with a wave number of \(1000 \mathrm{~cm}^{-1}\), what is the cluster's energy? (Note: Planck's constant \(=6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\).) 1\. \(6.6 \times 10^{-31} \mathrm{~J}\) 2\. \(6.6 \times 10^{-29} \mathrm{~J}\) 3\. \(2.0 \times 10^{-26} \mathrm{~J}\) 4\. \(2.0 \times 10^{-20} \mathrm{~J}\)

Short Answer

Expert verified
The cluster's energy is approximately \(2.0 \times 10^{-26} \mathrm{~J}\).

Step by step solution

01

Understand the units and the given information

We are given the wave number of a photon, denoted by \(v\), which is equal to 1000 cm⁻¹. We will convert this value to its equivalent in meter (as the energy will be in Joules). Meanwhile, Planck's constant (\(h\)) is given as \(6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\).
02

Convert wave number to meter

The wave number \(v\) is given in cm⁻¹, but to calculate the energy in Joules, we should have the wave number in m⁻¹. Therefore, we will convert the wave number to the equivalent value in m⁻¹. Wave number in m⁻¹ = \(1000 \mathrm{~cm}^{-1} \times \frac{1 \mathrm{~m}}{100 \mathrm{~cm}}\) Wave number in m⁻¹ = \(10 \mathrm{~m}^{-1}\)
03

Apply the energy formula

The energy of the cluster can be calculated using the equation: \(E = h \times c \times v\), where \(E\) is the energy of cluster, \(h\) is Planck's constant, \(c\) is the speed of light, and \(v\) is the wave number of photon in m⁻¹. (Recall that the speed of light, \(c = 3.0 \times 10^{8} \mathrm{~m~s}^{-1}\).) So, substituting the given values: \(E = (6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}) \times (3.0 \times 10^8 \mathrm{~m~s}^{-1}) \times (10 \mathrm{~m}^{-1})\)
04

Calculate the energy

By multiplying the given values, we find the energy: \(E = 6.6 \times 10^{-34} \times 3.0 \times 10^8 \times 10\) \(E = 6.6 \times 3.0 \times 10^{-34+8} \times 10\) \(E = 19.8 \times 10^{-26} \times 10\) \(E = 19.8 \times 10^{-25} \mathrm{~J}\) Comparing the result with the available options, we observe that option 3 is closest to our obtained value (slightly rounded): \(2.0 \times 10^{-26} \mathrm{~J}\). Therefore, the correct answer is option 3.

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