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Chapter 8: Problem 9

Use the Balmer equation (8.2) to determine (a) the frequency, in \(s^{-1}\), of the radiation corresponding to \(n=5\) (b) the wavelength, in nanometers, of the line in the Balmer series corresponding to \(n=7\) (c) the value of \(n\) corresponding to the Balmer series line at \(380 \mathrm{nm}\)

Short Answer

Expert verified
The calculated frequency for \(n=5\) is about \(6.166 \times 10^{14} s^{-1}\), the wavelength for \(n=7\) is approximately \(397 nm\) and the value of \(n\) for the Balmer series line at 380 nm is approximately 4.

Step by step solution

01

Calculation of frequency

Use the Balmer equation to find the wavelength for \( n=5 \), then use the formula \( v = \frac{c}{\lambda} \) to find the frequency. \( \lambda = \frac{1}{R_H (\frac{1}{2^2} - \frac{1}{5^2})} \) and \( v = \frac{c}{\lambda} \) Compute the values to get the frequency in \( s^{-1} \).
02

Calculation of wavelength

Again use the Balmer equation, this time to find the wavelength for \( n=7 \). Substitute the values to compute the wavelength in meters, and then convert to nanometers by multiplying the result by \( 10^9 \). That is, \( \lambda = \frac{1}{R_H (\frac{1}{2^2} - \frac{1}{7^2})} \times 10^9 nm \). Calculate this to get the wavelength in nanometers.
03

Finding the value of n

Rearrange the Balmer equation to solve for n. \( n = \sqrt{\frac{1}{(\frac{1}{\lambda} - \frac{1}{4R_H})}} \). Substitute the given wavelength at 380 nm (converting to meters by multiplying \( 10^{-9} \)) into the equation and solve to get the value of n.

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Most popular questions from this chapter

What electron transition in a hydrogen atom, starting from the orbit \(n=7,\) will produce light of wavelength \(410 \mathrm{nm} ?\)

The greatest probability of finding the electron in a small-volume element of the 1 s orbital of the hydrogen atom is at the nucleus. Yet the most probable distance of the electron from the nucleus is \(53 \mathrm{pm}\). How can you reconcile these two statements?

In everyday usage, the term "quantum jump" describes a change of a very significant magnitude compared to more gradual, incremental changes; it is similar in meaning to the term "a sea change." Does quantum jump have the same meaning when applied to events at the atomic or molecular level? Explain.

Which of the following statements is (are) correct for an electron with \(n=4\) and \(m_{\ell}=2 ?\) Explain. (a) The electron is in the fourth principal shell. (b) The electron may be in a \(d\) orbital. (c) The electron may be in a \(p\) orbital. (d) The electron must have \(m_{s}=+\frac{1}{2}\)

Write an acceptable value for each of the missing quantum numbers. (a) \(n=3, \ell=?, m_{\ell}=2, m_{s}=+\frac{1}{2}\) (b) \(n=?, \ell=2, m_{\ell}=1, m_{s}=-\frac{1}{2}\) (c) \(n=4, \ell=2, m_{\ell}=0, m_{s}=?\) (d) \(n=?, \ell=0, m_{\ell}=?, m_{s}=?\)
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