For the Lyman series, $$ {n_f} = 1 $$ . how can we calculate the $$ {E_{photon}} $$ for the bandhead of the Lyman series (the transition $$ n = \infty \to n = 1 $$ for emission) in joules and in eV?

Hint: The Lyman series is present in the ultraviolet region. To approach this question we need to use the Rydberg formula. The atomic hydrogen tends to display the emission spectrum. This spectrum has the tendency to enfolds the several spectral series. The Rydberg formula helps in relating the energy difference between the various levels of the Bohr’s model and the wavelength that has absorbed or emitted the photons.

Complete step by step answer:
So the Rydberg formula which we will using is the following:
$$ \dfrac{1}{\lambda } = R(\dfrac{1}{{{n^2}_f}} - \dfrac{1}{{{n_i}^2}}) $$
$$ \lambda $$ The represents the wavelength of the photon that has been emitted.
R represents the Rydberg constant which is equal to $$ 1.097 \times {10^7}{m^{ - 1}} $$ .
$$ {n_f} $$ Represents the final energy level of the transition
$$ {n_i} $$ Represents the initial energy level of the transition
In the question we have : $$ {n_i} = \infty \to {n_f} = 1 $$
So this transition is a part of the Lyman series. So in this case we have $$ {n_i} = \infty $$ which means
$$ \dfrac{1}{{{n_i}^2}} = 0 $$
So now substituting the values in formula we get,
$$ \dfrac{1}{\lambda } = R(\dfrac{1}{{{1^2}}} - 0) $$
On simplifying we get,
$$ \dfrac{1}{\lambda } = R $$
So,
$$ \dfrac{1}{R} = \lambda $$
On substituting the value we get,
$$ \lambda = \dfrac{1}{{1.097 \times {{10}^7}}} $$
$$ \lambda = 9.158 \times {10^{ - 8}}m $$
Now we will use the Planck Einstein relation to find the energy of the photon. The formula is:
$$ E = \dfrac{{hc}}{\lambda } $$
Here E is the energy of photon
The h is the Planck constant which is equal to $$ 6.626 \times {10^{ - 34}}Js $$ .
The c is the speed of light equal to $$ 3 \times {10^8}m{s^{ - 1}} $$
On substituting the values we get,
$$ E = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{9.158 \times {{10}^{ - 8}}}} = 2.17 \times {10^{ - 18}}J $$
We have calculated the answer in joules.
Now we will calculate it in eV. So ,
$$ 1eV = 1.6 \times {10^{ - 19}}J $$
So $$ 2.17 \times {10^{ - 18}}J $$ will be equal to= $$ 2.17 \times {10^{ - 18}}J \times \dfrac{{1eV}}{{1.6 \times {{10}^{ - 19}}J}} = 13.6eV $$
This means that we need 13.6eV to ionize the hydrogen atom.

Note: The Lyman series was discovered in the years between 1906- 1914 by Theodore Lyman. So according to Bohr's model the series of Lyman is displayed when the transition of the electron takes place from the higher energy state to the $${n_1} = 1$$ energy state. The wavelength of the Lyman series comes under the ultraviolet band.