A hydrogen atom in its fourth excited state emits a photon with a wavelength of 1282 nm. What is the atom's maximum possible orbital angular momentum (as a multiple of ℏ ) after the emission?

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Identify the initial state of the hydrogen atom. The fourth excited state corresponds to the principal quantum number \( n = 5 \), since the ground state is \( n = 1 \).

Determine the final state of the hydrogen atom after the photon emission. Use the wavelength of the emitted photon (1282 nm) and the energy difference formula for hydrogen: \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. This energy corresponds to the difference between the initial and final energy levels: \( E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \), where \( n_i = 5 \) and \( n_f \) is the final principal quantum number.

Solve for \( n_f \) (the final principal quantum number) using the energy difference equation. This will give the final state of the hydrogen atom after the photon emission.

Recall that the maximum possible orbital angular momentum of the atom is determined by the quantum number \( l \), which can take values from \( 0 \) to \( n_f - 1 \). The maximum orbital angular momentum corresponds to \( l = n_f - 1 \).

Express the maximum orbital angular momentum as a multiple of \( \hbar \) using the formula \( L = \sqrt{l(l+1)} \hbar \). Substitute \( l = n_f - 1 \) into this formula to find the maximum possible orbital angular momentum after the emission.

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Quantum States and Energy Levels

In quantum mechanics, atoms exist in discrete energy levels or states. The hydrogen atom's energy levels are quantized, meaning electrons can only occupy specific orbits defined by quantum numbers. The fourth excited state corresponds to the principal quantum number n=4, indicating the electron is in a higher energy level compared to the ground state.

Photon Emission and Wavelength

When an electron transitions from a higher energy level to a lower one, it emits a photon, which is a particle of light. The wavelength of the emitted photon is inversely related to the energy difference between the two states, as described by the equation E = hc/λ, where E is energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. In this case, the emitted photon has a wavelength of 1282 nm.

Orbital Angular Momentum

The orbital angular momentum of an electron in an atom is quantized and is given by the formula L = √(l(l+1))ℏ, where l is the azimuthal quantum number and ℏ is the reduced Planck's constant. For a hydrogen atom, the maximum possible value of l for the fourth excited state (n=4) is 3, leading to a maximum orbital angular momentum of 3ℏ. This concept is crucial for understanding the behavior of electrons in atomic orbitals.