Question

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

Answer 1

2

Answer 2

1. For a hydrogen atom, the energy is given by

   $\displaystyle E_n = -\frac{13.6}{n^2}$ eV.

2. The fourth excited state corresponds to $n_i = 5$ (since ground state $n=1$).

3. Energy of electron in initial state:

   $\displaystyle E_5 = -\frac{13.6}{25} \approx -0.544$ eV.

4. Let the final state be $n_f = n$. The energy emitted during the transition is:

   $\displaystyle E_\gamma = E_5 - E_n = 2.86$ eV.

5. So,

   $\displaystyle E_n = E_5 - 2.86 \approx -0.544 - 2.86 = -3.404$ eV.

6. Equate with the energy level expression:

   $-\frac{13.6}{n^2} = -3.404$

   $\frac{13.6}{n^2} = 3.404$

   $n^2 = \frac{13.6}{3.404} = 4$

   $n = 2$.

Use $E_n = -\frac{13.6}{n^2}$ and compute the energy difference $E_5 - E_n = 2.86$ eV. Solve for $n$.