Answer
$\sqrt{6} \;\hbar$
Work Step by Step
To solve this problem, we need to determine the maximum possible orbital angular momentum of the hydrogen atom after it emits a photon.
The problem states that the hydrogen atom is in its fourth excited state. This means the principal quantum number is 5 since the ground state is $ n = 1 $, and the fourth excited state is $ n = 5 $.
$$
n_i = 5
$$
where $i$ refers to the initial state.
The energy of the emitted photon is given by
$$
E_{\text{photon}} = \frac{hc}{\lambda}
$$
Plug the known:
$$
E_{\text{photon}} = \frac{(6.626 \times 10^{-34}) \times (3.0 \times 10^8)}{1282 \times 10^{-9}} =\bf 1.55 \times 10^{-19}\;\rm J
$$
$$
E_{\text{photon}} = \bf 0.97\rm \, \text{eV}
$$
The energy of the hydrogen atom at an energy level $ n $ is given by:
$$
E_n = - \frac{13.6}{n^2}
$$
For $ n_{\text{i}} = 5 $:
$$
E_5 = - \frac{13.6}{5^2} = - \frac{13.6}{25} =\bf -0.544 \, \rm {eV}
$$
The energy difference between the initial and final states corresponds to the energy of the photon. Using the photon energy $ 0.97 \, \text{eV} $, we can find the final energy level $ n_{\text{f}} $ such that:
$$
E_{\text{photon}} = E_i - E_{f}
$$
$$
E_{f} = E_{i} -E_{\text{photon}}
$$
Plug the known;
$$
E_f= -0.544 -0.97 =\bf -1.514 \, \rm {eV}
$$
Now solve for $ n_f$ using the energy formula for hydrogen levels:
$$
E_f = - \frac{13.6}{n_f^2}
$$
$$
n_f =\sqrt{ - \frac{13.6}{E_f }}
$$
Plug the known;
$$
n_f =\sqrt{ - \frac{13.6}{-1.514}}= \bf 2.997\approx 3
$$
The maximum possible orbital angular momentum $ L $ is given by
$$
L = \sqrt{l(l+1)} \hbar
$$
where $ l $ is the orbital angular momentum quantum number. The maximum $ l $ is $ l = n-1 $, so for $ n_f= 3 $:
$$
l_{\text{max}} = 3 - 1 = 2
$$
So,
$$
L = \sqrt{2(2+1)} \hbar = \color{red}{\sqrt{6} \;\hbar}
$$
