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Chapter 7: Problem 32

Calculate the frequency (Hz) and wavelength (nm) of the emitted photon when an electron drops from the \(n=4\) to the \(n=2\) level in a hydrogen atom.

Short Answer

Expert verified
The frequency of the emitted photon is \(6.15 * 10^{14} Hz\) and the wavelength is \(487 nm\).

Step by step solution

01

Calculate the energy difference

First, calculate the energy difference using the formula for energy levels in a hydrogen atom, \[ -13.6 eV \cdot \left(1 - \frac{1}{4}\right)\]. The energy of the higher level \(n=4\) is \(-13.6/4^2 = -0.85 eV\), for the lower level \(n=2\) it is \(-13.6/2^2 = -3.4 eV\). So, the energy difference is \(-0.85 eV - (-3.4 eV) = 2.55 eV\). To convert the energy from electron volts to Joules, which is a more common unit in such calculations, use the conversion \(1 eV = 1.6*10^{-19} J\). So, the energy difference is \(2.55 eV \cdot 1.6*10^{-19} J/eV = 4.08*10^{-19} J\).
02

Calculate frequency

The frequency can be found now using the equation \(E = h \cdot f\). Here, Planck's constant \(h = 6.63*10^{-34} Js\). Rearranging for \(f\), we get \(f = \frac{E}{h} = \frac{4.08*10^{-19} J}{6.63*10^{-34} Js} = 6.15 * 10^{14} Hz\).
03

Calculate the wavelength

Finally, to calculate the wavelength, use the wave equation \(c = f \cdot \lambda\), where \(c = 3 * 10^{8} ms^{-1}\), the speed of light. Rearranging for \(\lambda\), we get \(\lambda = \frac{c}{f} = \frac{3 * 10^{8} ms^{-1}}{6.15*10^{14} Hz} = 4.87 * 10^{-7} m\). To convert the result to nanometers, a common unit in this context, multiply by \(10^{9} nm/m\). So, \(\lambda = 4.87 * 10^{-7} m \cdot 10^{9} nm/m = 487 nm\).

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Most popular questions from this chapter

Some copper compounds emit green light when they are heated in a flame. How would you determine whether the light is of one wavelength or a mixture of two or more wavelengths?

An electron in an excited state in a hydrogen atom can return to the ground state in two different ways: (a) via a direct transition in which a photon of wavelength \(\lambda_{1}\) is emitted and (b) via an intermediate excited state reached by the emission of a photon of wavelength \(\lambda_{2}\). This intermediate excited state then decays to the ground state by emitting another photon of wavelength \(\lambda_{3}\). Derive an equation that relates \(\lambda_{1}\) to \(\lambda_{2}\) and \(\lambda_{3}\)

What are emission spectra? How do line spectra differ from continuous spectra?

Describe the shapes of \(s, p,\) and \(d\) orbitals. How are these orbitals related to the quantum numbers \(n, \ell\) and \(m_{\ell} ?\)

What is the difference between a \(2 p_{x}\) and a \(2 p_{y}\) orbital?
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