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.
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.
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Correct Answer: 2
Approach Solution - 1
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.
Approach Solution -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 \( n = 2 \).
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