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Chapter 7: Problem 47

Microwave radiation has a wavelength on the order of \(1.0 \mathrm{~cm} .\) Calculate the frequency and the energy of a single photon of this radiation. Calculate the energy of an Avogadro's number of photons (called an \(einstein\)) of this radiation.

Short Answer

Expert verified
The frequency of the microwave radiation is \(3.0 \times 10^{10} Hz\). The energy of a single photon of this microwave radiation is \(1.988 \times 10^{-23} J\). The energy of an einstein of this microwave radiation is approximately \(11.972 J\).

Step by step solution

01

Find the frequency of the microwave radiation

We are given the wavelength of the microwave radiation as \(1.0 cm\). We can find the frequency using the electromagnetic wave relationship: \(c = \lambda * \nu \) where: - \(c\) is the speed of light (\(3.0 \times 10^8 m/s\)) - \(\lambda\) is the wavelength of the radiation (1.0 cm) - \(\nu\) is the frequency of the radiation First, we need to convert the wavelength from centimeters to meters: \(1.0 cm = \frac{1}{100} m\) Next, we'll rearrange the equation to solve for the frequency: \(\nu = \frac{c}{\lambda}\)
02

Calculate the frequency

Plug in the values for c and the converted wavelength: \(\nu = \frac{3.0 \times 10^8 m/s}{\frac{1}{100} m}\) \(\nu = 3.0 \times 10^{10} Hz\) So, the frequency of the microwave radiation is \(3.0 \times 10^{10} Hz\).
03

Calculate the energy of a single photon

To find the energy of a single photon, we'll use Planck's equation: \(E = h * \nu\) where \(h\) is Planck's constant (\(6.626 \times 10^{-34} Js\)) and \(\nu\) is the frequency of the microwave radiation. Plug in the values for the frequency and Planck's constant: \(E = (6.626 \times 10^{-34} Js) * (3.0 \times 10^{10} Hz)\) \(E = 1.988 \times 10^{-23} J\) So, the energy of a single photon of this microwave radiation is \(1.988 \times 10^{-23} J\).
04

Calculate the energy of Avogadro's number of photons (einstein)

To find the energy of an einstein of this microwave radiation, we'll multiply the energy of a single photon by Avogadro's number: \(E_einstein = E_{photon} * N_A\) where: - \(E_einstein\) is the energy of an einstein of microwave radiation - \(E_{photon}\) is the energy of a single photon from step 3 (1.988 x 10^(-23) J) - \(N_A\) is Avogadro's number (\(6.022\times 10^{23} mol^{-1}\)) Plug in the values: \(E_einstein = (1.988 \times 10^{-23} J) * (6.022\times 10^{23} mol^{-1})\) \(E_einstein = 1.1972 \times 10^1 J\) So, the energy of an einstein of this microwave radiation is approximately \(11.972 J\).

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Most popular questions from this chapter

One of the visible lines in the hydrogen emission spectrum corresponds to the \(n=6\) to \(n=2\) electronic transition. What color light is this transition? See Exercise 150 .

Which of the following sets of quantum numbers are not allowed in the hydrogen atom? For the sets of quantum numbers that are incorrect, state what is wrong in each set. a. \(n=3, \ell=2, m_{\ell}=2\) b. \(n=4, \ell=3, m_{\ell}=4\) c. \(n=0, \ell=0, m_{\ell}=0\) d. \(n=2, \ell=-1, m_{\ell}=1\)

Answer the following questions based on the given electron configurations, and identify the elements. a. Arrange these atoms in order of increasing size: $[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{6} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{1} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{3}$ b. Arrange these atoms in order of decreasing first ionization energy: [Ne $3 s^{2} 3 p^{5} ;[\operatorname{Ar}] 4 s^{2} 3 d^{10} 4 p^{3} ;[\operatorname{Ar}] 4 s^{2} 3 d^{10} 4 p^{5}$

A certain oxygen atom has the electron configuration 1$s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} .$ How many unpaired electrons are present? Is this an excited state of oxygen? In going from this state to the ground state, would energy be released or absorbed?

What are the possible values for the quantum numbers \(n, \ell,\) and $m_{\ell} ?$
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