To find the integer value of $ n $ for which an electron transitions from the fourth excited state in a hydrogen atom, thus emitting a photon with an energy of 2.86 eV, we can follow these steps:

We use the formula for the energy difference associated with electron transitions in a hydrogen atom:

$ E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $

However, in this question, it seems there's a typo in the original explanation, as both indices are given as $\mathrm{n}_1$. To correct it, we should use this formula:

$ E = 13.6 \left( \frac{1}{n^2} - \frac{1}{5^2} \right) $

where $ n_1 = 5 $ (the fifth energy level or fourth excited state) and $ n_2 = n $ (the state to which the electron transitions).

Given the photon's energy is 2.86 eV, set up the equation:

$ 2.86 = 13.6 \left( \frac{1}{n^2} - \frac{1}{25} \right) $

Solve for $\frac{1}{n^2}$:

$ \frac{1}{n^2} = \frac{2.86}{13.6} + \frac{1}{25} $

Calculate the value:

$ \frac{1}{n^2} = 0.21 + 0.04 = 0.25 $

Find $ n^2 $:

$ n^2 = \frac{1}{0.25} = 4 $

Consequently:

$ n = \sqrt{4} = 2 $

Thus, the electron transitions to the $ n = 2 $ energy state.