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Chapter 7: Problem 165

Assume that a hydrogen atom's electron has been excited to the \(n=6\) level. How many different wavelengths of light can be emitted as this excited atom loses energy?

Short Answer

Expert verified
The electron in the hydrogen atom is excited to the \(n=6\) level and can fall to any level below the current level (n=1, 2, 3, 4, or 5). Each transition corresponds to a different energy and, consequently, a different wavelength of emitted light. Counting all the unique transitions, we find that the hydrogen atom can emit 5 different wavelengths of light as it loses energy.

Step by step solution

01

Identify the energy levels involved in the transitions

The electron in the hydrogen atom is excited to the n=6 level, and it can fall to any level below the current level (n=1, 2, 3, 4, or 5). Each transition corresponds to a different energy, and consequently, a different wavelength of emitted light.
02

Calculate the number of possible transitions (wavelengths)

When the electron falls from the n=6 level to a lower level, we can calculate the total number of possible transitions by adding up the possible transitions from each lower level to n=6. Transitions to n=1: 1. n=6 → n=1 Transitions to n=2: 2. n=6 → n=2 Transitions to n=3: 3. n=6 → n=3 Transitions to n=4: 4. n=6 → n=4 Transitions to n=5: 5. n=6 → n=5
03

Count all the unique transitions

We can now count all the unique transitions that we listed above, which will give us the number of different wavelengths of light emitted as the excited hydrogen atom loses energy. Number of unique transitions = 5 Therefore, the hydrogen atom with an electron excited to the n=6 level can emit 5 different wavelengths of light as it loses energy.

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Most popular questions from this chapter

In defining the sizes of orbitals, why must we use an arbitrary value, such as 90% of the probability of finding an electron in that region?

One of the emission spectral lines for \(\mathrm{Be}^{3+}\) has a wavelength of 253.4 \(\mathrm{nm}\) for an electronic transition that begins in the state with \(n=5 .\) What is the principal quantum number of the lower-energy state corresponding to this emission? (Hint: The Bohr model can be applied to one- electron ions. Don't forget the \(Z\) factor: \(Z=\) nuclear charge \(=\) atomic number.)

The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and 589.6 nm. What are the frequency and the energy of a photon of light at each of these wavelengths? What are the energies in kJ/mol?

Are the following statements true for the hydrogen atom only, true for all atoms, or not true for any atoms? a. The principal quantum number completely determines the energy of a given electron. b. The angular momentum quantum number, \(\ell,\) determines the shapes of the atomic orbitals. c. The magnetic quantum number, \(m_{\ell},\) determines the direction that the atomic orbitals point in space.

The Heisenberg uncertainty principle can be expressed in the form $$\Delta E \cdot \Delta t \geq \frac{h}{4 \pi}$$ where \(E\) represents energy and \(t\) represents time. Show that the units for this form are the same as the units for the form used in this chapter: $$\Delta x \cdot \Delta(m v) \geq \frac{h}{4 \pi}$$
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