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Chapter 6: Problem 46

Place the following transitions of the hydrogen atom in order from shortest to longest wavelength of the photon emitted: \(n=5\) to \(n=3, n=4\) to \(n=2, n=7\) to \(n=4,\) and \(n=3\) to \(n=2\) .

Short Answer

Expert verified
The order of the given transitions from shortest to longest wavelength of the photon emitted is: \(n=5\) to \(n=3\), \(n=7\) to \(n=4\), \(n=4\) to \(n=2\), and \(n=3\) to \(n=2\).

Step by step solution

01

Understanding the Rydberg formula for hydrogen atom

The Rydberg formula can be used to find the wavelength of the emitted photon during an electron transition in the hydrogen atom. The formula is given by: \[ \frac{1}{\lambda} = R_{H} \left( \frac{1}{n_{1}^2} - \frac{1}{n_{2}^2} \right) \] where \(\lambda\) is the wavelength, \(R_{H}\) is the Rydberg constant for hydrogen atom (\(1.097 \times 10^7 m^{-1}\)), \(n_{1}\) is the lower energy level, and \(n_{2}\) is the higher energy level.
02

Calculate the wavelength for each transition

We'll plug in the given values for \(n_1\) and \(n_2\) in the Rydberg formula to find the wavelength for each transition: 1) \(n=5 \rightarrow n=3\): \[ \frac{1}{\lambda_{1}} = R_H\left(\frac{1}{3^2} - \frac{1}{5^2}\right) \] 2) \(n=4 \rightarrow n=2\): \[ \frac{1}{\lambda_{2}} = R_H\left(\frac{1}{2^2} - \frac{1}{4^2}\right) \] 3) \(n=7 \rightarrow n=4\): \[ \frac{1}{\lambda_{3}} = R_H\left(\frac{1}{4^2} - \frac{1}{7^2}\right) \] 4) \(n=3 \rightarrow n=2\): \[ \frac{1}{\lambda_{4}} = R_H\left(\frac{1}{2^2} - \frac{1}{3^2}\right) \]
03

Solving for the wavelengths

Calculate the value of \(\lambda_i\) for each transition: 1) \(\lambda_{1} = 1.6 \times 10^{-7}\,m\) 2) \(\lambda_{2} = 4.86 \times 10^{-7}\,m\) 3) \(\lambda_{3} = 2.97 \times 10^{-7}\,m\) 4) \(\lambda_{4} = 6.56 \times 10^{-7}\,m\)
04

Arrange the wavelengths from shortest to longest

With the calculated values for each wavelength, we can now arrange them in the required order: 1) \(1.6 \times 10^{-7}\,m\) (\(n=5 \rightarrow n=3\)) 2) \(2.97\times 10^{-7}\,m\) (\(n=7 \rightarrow n=4\)) 3) \(4.86 \times 10^{-7}\,m\) (\(n=4 \rightarrow n=2\)) 4) \(6.56 \times 10^{-7}\,m\) (\(n=3 \rightarrow n=2\)) So, the order of the given transitions from shortest to longest wavelength of the photon emitted is: \(n=5\) to \(n=3\), \(n=7\) to \(n=4\), \(n=4\) to \(n=2\), and \(n=3\) to \(n=2\).

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Most popular questions from this chapter

Indicate whether energy is emitted or absorbed when the following electronic transitions occur in hydrogen: (a) from \(n=2\) to \(n=6,(\mathbf{b})\) from an orbit of radius 4.76\(\hat{\mathrm{A}}\) to one of radius \(0.529 \mathrm{A},(\mathbf{c})\) from the \(n=6\) to the \(n=9\) state.

Classify each of the following statements as either true or false: (a) A hydrogen atom in the \(n=3\) state can emit light at only two specific wavelengths, (b) a hydrogen atom in the \(n=2\) state is at a lower energy than one in the \(n=1\) state, and (c) the energy of an emitted photon equals the energy difference of the two states involved in the emission.

How many unique combinations of the quantum numbers \(l\) and \(m_{l}\) are there when (a) \(n=3,\) (b) \(n=4 ?\)

(a) For an He^ + ion, do the 2 s and 2\(p\) orbitals have the same energy? If not, which orbital has a lower energy? (b) If we add one electron to form the He atom, would your answer to part (a) change?

The Lyman series of emission lines of the hydrogen atom are those for which \(n_{1}=1 .\) (a) Determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed. (b) Calculate the wavelengths of the first three lines in the Lyman series - those for which \(n_{1}=2,3,\) and \(4 .\)
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