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

This problem requires finding the final principal quantum number (\(n\)) of an electron in a hydrogen atom after it transitions from a higher energy state by emitting a photon of a specific energy.

Concept Used:

The energy of an electron in the \(k^{th}\) energy level (or orbit) of a hydrogen atom is given by the formula:

\[ E_k = -\frac{13.6}{k^2} \text{ eV} \]

The "ground state" corresponds to \(k=1\). The "first excited state" corresponds to \(k=2\), the "second excited state" to \(k=3\), and so on. Therefore, the \(m^{th}\) excited state corresponds to the principal quantum number \(k = m+1\).

When an electron makes a transition from a higher initial energy state \(E_i\) (with quantum number \(k_i\)) to a lower final energy state \(E_f\) (with quantum number \(k_f\)), it emits a photon. The energy of this photon (\(\Delta E\)) is equal to the difference in the energy levels.

\[ \Delta E = E_i - E_f \]

Step-by-Step Solution:

Step 1: Determine the initial principal quantum number (\(k_i\)) of the electron.

The problem states that the electron is initially in the "fourth excited state". Using the relationship that the \(m^{th}\) excited state corresponds to \(k = m+1\), for the fourth excited state (\(m=4\)), the principal quantum number is:

\[ k_i = 4 + 1 = 5 \]

Step 2: Calculate the energy of the electron in this initial state (\(E_i\)).

Using the energy formula with \(k_i = 5\):

\[ E_i = E_5 = -\frac{13.6}{5^2} \text{ eV} = -\frac{13.6}{25} \text{ eV} \] \[ E_i = -0.544 \text{ eV} \]

Step 3: Use the energy of the emitted photon to find the energy of the final state (\(E_n\)).

The energy of the emitted photon is given as \(\Delta E = 2.86\) eV. The final state is the \(n^{th}\) energy state, so its energy is \(E_n\). The relationship is:

\[ \Delta E = E_i - E_n \]

Rearranging to solve for \(E_n\):

\[ E_n = E_i - \Delta E \]

Substituting the known values:

\[ E_n = -0.544 \text{ eV} - 2.86 \text{ eV} \] \[ E_n = -3.404 \text{ eV} \]

Step 4: Determine the principal quantum number (\(n\)) of the final energy state.

We use the energy formula again, this time with the final energy \(E_n\):

\[ E_n = -\frac{13.6}{n^2} \text{ eV} \]

Substituting the value of \(E_n\) we just found:

\[ -3.404 = -\frac{13.6}{n^2} \]

Now, we solve for \(n^2\):

\[ n^2 = \frac{13.6}{3.404} \] \[ n^2 \approx 3.995 \]

Since the principal quantum number \(n\) must be an integer, we can round the value of \(n^2\) to the nearest integer.

\[ n^2 = 4 \] \[ n = \sqrt{4} = 2 \]

Thus, the integer value of n for the final energy state is 2.

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 \( n = 2 \).