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Chapter 38: Problem 34

A hydrogen atom is in its fifth excited state, with principal quantum number \(n=6 .\) The atom emits a photon with a wavelength of \(410 \mathrm{nm}\). Determine the maximum possible orbital angular momentum of the electron after emission.

Short Answer

Expert verified
Answer: The maximum possible orbital angular momentum of the electron after the photon emission is approximately \(1.49 \times 10^{-34} \mathrm{Js}\).

Step by step solution

01

Calculate the energy of the emitted photon.

The energy of a photon can be calculated using its wavelength and the Planck's constant, with the formula \(E = \frac{hc}{\lambda}\), where \(E\) is the energy, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), \(c\) is the speed of light (\(2.998 \times 10^{8} \mathrm{m/s}\)), and \(\lambda\) is the wavelength of the photon (\(410 \mathrm{nm}\)). Converting the wavelength to meters, we get \(\lambda = 410 \times 10^{-9} \mathrm{m}\). Now, we can calculate the energy of the photon: \(E = \frac{(6.626 \times 10^{-34} \mathrm{Js})(2.998 \times 10^{8} \mathrm{m/s})}{410 \times 10^{-9} \mathrm{m}} \approx 4.839 \times 10^{-19} \mathrm{J}\).
02

Calculate the initial energy level of the hydrogen atom.

The energy of a hydrogen atom with a principal quantum number \(n\) is given by the formula \(E_n = -\frac{13.6 \mathrm{eV}}{n^2}\). Since the atom is initially in its fifth excited state (with \(n = 6\)), its initial energy can be calculated as: \(E_6 = -\frac{13.6 \mathrm{eV}}{6^2} \approx -0.377 \mathrm{eV}\). Converting the energy to joules, we get \(E_6 \approx -6.044 \times 10^{-20} \mathrm{J}\).
03

Calculate the final energy level of the hydrogen atom.

Since the energy of a photon is equal to the difference in energy levels between the initial and final states, we can calculate the final energy level (\(E_f\)) after photon emission as follows: \(E_f = E_6 - E_{photon} \approx -6.044 \times 10^{-20} \mathrm{J} - 4.839 \times 10^{-19} \mathrm{J} \approx -5.443 \times 10^{-19} \mathrm{J}\). Converting the energy back to electron volts, we get \(E_f \approx -3.39 \mathrm{eV}\).
04

Determine the maximum value of the azimuthal quantum number (l).

To find the final principal quantum number \(n_f\), we can compare the obtained energy to the energy levels of the hydrogen atom: \(E_{n_f} = -\frac{13.6 \mathrm{eV}}{n_f^2}\). Setting \(E_{n_f} = -3.39 \mathrm{eV}\), we can solve for \(n_f\): \(n_f \approx \sqrt{\frac{13.6 \mathrm{eV}}{3.39 \mathrm{eV}}} \approx 2\). The maximum possible value of the azimuthal quantum number (\(l\)) is \(n_f - 1\) for a given energy level. Therefore, the maximum value of \(l\) for the final state is: \(l_{max} = n_f - 1 = 2 - 1 = 1\).
05

Calculate the maximum possible orbital angular momentum.

The orbital angular momentum (\(L\)) can be calculated using the formula \(L = \sqrt{l(l+1)}\hbar\), where \(\hbar\) is the reduced Planck's constant (\(1.054 \times 10^{-34} \mathrm{Js}\)). Using the maximum value of \(l\), we can calculate the maximum possible orbital angular momentum: \(L = \sqrt{1(1+1)}(1.054 \times 10^{-34} \mathrm{Js}) \approx 1.49 \times 10^{-34} \mathrm{Js}\). The maximum possible orbital angular momentum of the electron after the photon emission is approximately \(1.49 \times 10^{-34} \mathrm{Js}\).

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Most popular questions from this chapter

Find the energy difference between the ground state of hydrogen and deuterium (hydrogen with an extra neutron in the nucleus)

Given that the hydrogen atom has an infinite number of energy levels, why can't a hydrogen atom in the ground state absorb all possible wavelengths of light?

The radius of the \(n=1\) orbit in the hydrogen atom is $$ a_{0}=0.053 \mathrm{nm} $$ a) Compute the radius of the \(n=6\) orbit. How many times larger is this compared to the \(n=1\) radius? b) If an \(n=6\) electron relaxes to an \(n=1\) orbit (ground state), what is the frequency and wavelength of the emitted radiation? What kind of radiation was emitted (visible, infrared, etc.)? c) How would your answer in (a) change if the atom was a singly ionized helium atom \(\left(\mathrm{He}^{+}\right),\) instead?

Which of the following can be used to explain why you can't walk through walls? a) Coulomb repulsion d) the Pauli exclusion b) the strong nuclear force \(\quad\) principle c) gravity e) none of the above

An electron made a transition between allowed states emitting a photon. What physical constants are needed to calculate the energy of photon from the measured wavelength? (select all that apply) a) the Plank constant, \(h\) b) the basic electric charge, \(e\) c) the speed of light in vacuum, \(c\) d) the Stefan-Boltzmann constant, \(\sigma\)
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