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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

An electron in a hydrogen atom is in the ground state (1s). Calculate the probability of finding the electron within a Bohr radius \(\left(a_{0}=0.05295 \mathrm{nm}\right)\) of the proton. The ground state wave function for hydrogen is: \(\psi_{1 s}(r)=A_{1 s} e^{-r / a_{0}}=e^{-r / a_{0}} / \sqrt{\pi a_{0}^{3}}\).

The binding energy of an extra electron when As atoms are doped in a Si crystal may be approximately calculated by considering the Bohr model of a hydrogen atom. a) Show the ground energy of hydrogen-like atoms in terms of the dielectric constant and the ground state energy of a hydrogen atom. b) Calculate the binding energy of the extra electron in a Si crystal. (The dielectric constant of Si is about 10.0 , and the effective mass of extra electrons in a Si crystal is about \(20.0 \%\) of that of free electrons.)

A beam of electrons is incident upon a gas of hydrogen atoms. What minimum speed must the electrons have to cause the emission of light from the \(n=3\) to \(n=2\) transition of hydrogen?

In a hydrogen atom, the electron is in the \(n=5\) state. Which of the following sets could correspond to the \(\ell, m\) states of the electron? a) 5,-3 b) 4,-5 c) 3,-2 d) 4,-6

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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