An electron in the hydrogen atom initially in the fourth excited state makes a transition to $\mathrm{n}^{\text {th }}$ energy state by emitting a photon of energy 2.86 eV . The integer value of n will be__________.

<p>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:</p> <p><p>We use the formula for the energy difference associated with electron transitions in a hydrogen atom:</p> <p>$ E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $</p> <p>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:</p> <p>$ E = 13.6 \left( \frac{1}{n^2} - \frac{1}{5^2} \right) $</p> <p>where $ n_1 = 5 $ (the fifth energy level or fourth excited state) and $ n_2 = n $ (the state to which the electron transitions).</p></p> <p><p>Given the photon's energy is 2.86 eV, set up the equation:</p> <p>$ 2.86 = 13.6 \left( \frac{1}{n^2} - \frac{1}{25} \right) $</p></p> <p><p>Solve for $\frac{1}{n^2}$:</p> <p>$ \frac{1}{n^2} = \frac{2.86}{13.6} + \frac{1}{25} $</p></p> <p><p>Calculate the value:</p> <p>$ \frac{1}{n^2} = 0.21 + 0.04 = 0.25 $</p></p> <p><p>Find $ n^2 $:</p> <p>$ n^2 = \frac{1}{0.25} = 4 $</p></p> <p><p>Consequently:</p> <p>$ n = \sqrt{4} = 2 $</p></p> <p>Thus, the electron transitions to the $ n = 2 $ energy state.</p>