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Chapter 7: Problem 63

An excited hydrogen atom with an electron in the \(n=5\) state emits light having a frequency of \(6.90 \times 10^{14} \mathrm{~s}^{-1}\). Determine the principal quantum level for the final state in this electronic transition.

Short Answer

Expert verified
Using the given frequency of emitted light (\(v = 6.9 \times 10^{14} \mathrm{~s}^{-1}\)) and initial principal quantum number (n=5), we can use the Rydberg formula to determine the final principal quantum number (\(n_f\)) after the electronic transition. Plugging the values into the formula and solving for \(n_f\), we find that the final principal quantum number is approximately 3.

Step by step solution

01

Determine the initial principal quantum number

The problem states that the electron is initially in the \(n=5\) state. We'll assign this as \(n_i = 5\).
02

Recall the Rydberg formula

The Rydberg formula relates the frequency of light emitted during an electronic transition to the initial and final principal quantum numbers: \[v = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\] where \(v\) is the frequency of the emitted light, \(R_H\) is the Rydberg constant for hydrogen \((R_H = 3.29 \times 10^{15} \mathrm{~s}^{-1})\), \(n_i\) is the initial principal quantum number, and \(n_f\) is the final principal quantum number.
03

Insert the given values and solve for \(n_f\)

Insert the given values \(v = 6.9 \times 10^{14} \mathrm{~s}^{-1}\) and \(n_i = 5\) into the Rydberg formula and solve for \(n_f\): \[6.90 \times 10^{14} \mathrm{~s}^{-1} = 3.29 \times 10^{15} \mathrm{~s}^{-1} \left(\frac{1}{n_f^2} - \frac{1}{5^2}\right)\]
04

Rearrange the equation to isolate \(n_f\)

Rearrange the equation to isolate \(n_f^2\): \[n_f^2 = \frac{1}{\frac{1}{5^2} - \frac{6.90 \times 10^{14} \mathrm{~s}^{-1}}{3.29 \times 10^{15} \mathrm{~s}^{-1}}}\]
05

Calculate the final principal quantum number

Calculate the value of \(n_f\): \[n_f = \sqrt{\frac{1}{\frac{1}{5^2} - \frac{6.90 \times 10^{14} \mathrm{~s}^{-1}}{3.29 \times 10^{15} \mathrm{~s}^{-1}}}}\] \[n_f ≈ 3\] The final principal quantum number for the electronic transition is approximately 3.

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Most popular questions from this chapter

Consider the following ionization energies for aluminum: $$ \begin{aligned} \mathrm{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} & I_{1}=580 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} & I_{2}=1815 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ $$ \begin{array}{ll} \mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} & I_{3}=2740 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} & I_{4}=11,600 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\). c. Which one of the four ions has the greatest electron affinity? Explain. d. List the four aluminum ions given in order of increasing size, and explain your ordering. (Hint: Remember that most of the size of an atom or ion is due to its electrons.)

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