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Chapter 6: Problem 38
An electron is in the \(n=8\) level of ionized helium. (a) Find the threelongest wavelengths that are emitted when the electron makes a transition fromthe \(n=8\) level to a lower level. (b) Find the shortest wavelength that can beemitted. (c) Find the three longest wavelengths at which the electron in the\(n=8\) level will \(a b s o r b\) a photon and move to a higher state, if wecould somehow keep it in that level long enough to absorb. (d) Find the shortest wavelength that can be absorbed.
Short Answer
Expert verified
(a) n=7, 6, 5; (b) n=1; (c) n=9, 10, 11; (d) n=∞
Step by step solution
01
Understand the context
In this exercise, we are dealing with an electron in the n=8 level of ionized helium. Helium is a hydrogen-like atom with a nuclear charge of Z=2, meaning we need to use modified formulas for atomic transitions.
02
Identify the formula for wavelength in hydrogen-like atoms
The wavelength for transitions in hydrogen-like atoms is given by the Rydberg formula: \[\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\] where R is the Rydberg constant, Z is the atomic number (for helium, Z=2), \(n_1\) and \(n_2\) are the principal quantum numbers with \(n_2 > n_1\).
03
Calculate the longest wavelengths for emissions
For the three longest wavelengths, we need the transitions with the smallest energy difference, which means going from \(n=8\) to successively higher \(n_1\): \(n_1 = 7, 6, 5\). Compute:\(\frac{1}{\lambda} = R·Z^2\left(\frac{1}{n_1^2} - \frac{1}{8^2}\right)\), substituting R = 1.097 × 10^7 m^-1 and Z = 2.
04
Calculate the shortest emission wavelength
For the shortest wavelength, the transition with the largest energy difference is from \(n=8\) to \(n=1\). Compute:\(\frac{1}{\lambda} = R·Z^2\left(\frac{1}{1^2} - \frac{1}{8^2}\right)\).
05
Calculate the longest wavelengths for absorption
For the three longest wavelengths, the transitions involve moving to the next highest levels \(n_2 = 9, 10, 11\) from \(n=8\). Compute:\(\frac{1}{\lambda} = R·Z^2\left(\frac{1}{8^2} - \frac{1}{n_2^2}\right)\).
06
Calculate the shortest absorption wavelength
For the shortest wavelength, the transition goes to \(n=∞\). The formula simplifies to:\(\frac{1}{\lambda} = R·Z^2\left(\frac{1}{8^2} - 0\right)\).
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rydberg formula
The Rydberg formula is essential when dealing with electron transitions in hydrogen-like atoms, including ionized helium. This formula helps calculate the wavelengths of light associated with these transitions. It's given by: \[ \frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \]Where:
- R is the Rydberg constant, approximately 1.097 × 10^7 m^-1.
- Z is the atomic number (for helium, Z = 2).
- \(n_1\) and \(n_2\) are the energy levels (principal quantum numbers), with \(n_2\) always greater than \(n_1\) for emission and less for absorption.
Understanding this formula allows you to relate the energy levels of an electron to the wavelength of light emitted or absorbed. This is crucial for predicting the spectral lines of elements like helium.
Atomic energy levels
Atomic energy levels are the specific energies that an electron in an atom can have. Think of these as steps on a staircase; electrons can occupy these steps but not the spaces in between. For hydrogen-like atoms, these levels are given by the formula:\[ E_n = -\frac{Z^2 R h c}{n^2} \]Where:
- Z is the atomic number.
- R is the Rydberg constant.
- h is Planck's constant.
- c is the speed of light.
- n is the principal quantum number.
Each transition of an electron from a higher level (\(n_2\)) to a lower level (\(n_1\)) releases energy in the form of light, and the difference in energy corresponds to the photon’s wavelength. For effective calculations, it’s important to understand that the lower the quantum number, the more tightly the electron is bound to the nucleus and the higher its energy release upon transition.
Wavelength calculations
To calculate the wavelengths of the emitted or absorbed light, we use the Rydberg formula. When an electron transitions between energy levels, it either absorbs or emits a photon whose wavelength can be determined by:For emission: \[ \frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \]For absorption: \[ \frac{1}{\lambda} = RZ^2\left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right) \]The shortest wavelength corresponds to the largest energy difference, like transitioning from a high energy level to the lowest (for example, \(n=8\) to \(n=1\)). Meanwhile, the longest wavelengths result from transitions between levels with smaller energy differences (\(n=8\) to \(n=7\), \(n=6\), or \(n=5\) for example).By substituting the values of R, Z, and the appropriate quantum numbers into the formula, you can calculate the specific wavelengths for each transition.
Quantum numbers
Quantum numbers provide a way to describe the properties and locations of electrons within an atom. The principal quantum number (n) indicates the electron's energy level and distance from the nucleus. In our exercise with ionized helium, we primarily deal with n-values like n=8, n=7, and so on.The higher the value of n, the higher the energy level and the farther the electron is from the nucleus. Transitioning between these levels involves either absorption or emission of energy in the form of photons.Key points to remember:
- The principal quantum number (n) can be any positive integer.
- The energy gap between levels decreases as n increases.
- Quantum numbers are essential for applying the Rydberg formula and other calculations for transitions and wavelengths.
Understanding these numbers grants you deeper insight into atomic structures and the behavior of electrons during various transitions.
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