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 1cm.

Hint:

Use the formula for the energy levels of a hydrogen atom and the energy of the emitted photon during a transition between energy levels. Identify the initial and final energy levels based on the given information. Set up the equation relating the photon energy to the initial and final quantum numbers and solve for the unknown final quantum number \( n \).

Answer 1

Correct Answer: 2

The energy of an electron in the \( n^{th} \) orbit of a hydrogen atom is given by: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] The electron is initially in the fourth excited state. The ground state is \( n=1 \), the first excited state is \( n=2 \), the second excited state is \( n=3 \), the third excited state is \( n=4 \), and the fourth excited state is \( n=5 \). 
So, the initial energy level is \( n_i = 5 \). 
The electron makes a transition to the \( n^{th} \) energy state, so the final energy level is \( n_f = n \). 
The energy of the emitted photon is equal to the difference in energy between the initial and final energy levels: \[ E_{photon} = E_i - E_f = -\frac{13.6}{n_i^2} - \left( -\frac{13.6}{n_f^2} \right) = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \, \text{eV} \] Given that the energy of the emitted photon is 2.86 eV, and \( n_i = 5 \), we have: \[ 2.86 = 13.6 \left( \frac{1}{n^2} - \frac{1}{5^2} \right) \] \[ 2.86 = 13.6 \left( \frac{1}{n^2} - \frac{1}{25} \right) \] Divide both sides by 13.6: \[ \frac{2.86}{13.6} = \frac{1}{n^2} - \frac{1}{25} \] \[ 0.21029 \approx \frac{1}{n^2} - 0.04 \] \[ \frac{1}{n^2} = 0.21029 + 0.04 = 0.25029 \approx 0.25 = \frac{1}{4} \] \[ n^2 = 4 \] \[ n = \sqrt{4} = 2 \] Since \( n \) must be an integer, the final energy state is \( n = 2 \).