In a sample of H-atoms in ground state electrons make transition from ground state to a particular excited state where path length is 5 times de Broglie wavelength , electrons make back transition to the ground state producing all possible photons. If photon having 2nd highest energy of this sample can be used to excite the electron in a particular excited state of `Li^(2+)` ion then find the final excited state of `Li^(2+)` ion .

Answer

To solve the problem step by step, we will analyze the transitions of electrons in hydrogen and lithium ions, and calculate the final excited state of the Li^(2+) ion based on the energy of the photons emitted during the transition of the hydrogen atom.

### Step 1: Determine the excited state of the hydrogen atom
The problem states that the electron in a hydrogen atom transitions from the ground state to an excited state where the path length is 5 times the de Broglie wavelength.

The de Broglie wavelength (λ) is given by:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the electron.

For an electron moving in a circular orbit, the circumference of the orbit is given by:
\[ 2\pi r = n\lambda \]
where \(n\) is an integer (the principal quantum number).

Given that the path length is 5 times the de Broglie wavelength:
\[ 2\pi r = 5\lambda \]
We can set this equal to the equation for the circumference:
\[ n\lambda = 5\lambda \]
This implies:
\[ n = 5 \]
Thus, the electron in the hydrogen atom transitions to the \(n=5\) excited state.

### Step 2: Calculate the energy of the photon emitted during the transition
When the electron transitions back to the ground state (n=1), it emits a photon. The energy of the photon can be calculated using the formula:
\[ E = -13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Here, \(n_1 = 1\) (ground state) and \(n_2 = 5\) (excited state):
\[ E = -13.6 \, \text{eV} \left( \frac{1}{1^2} - \frac{1}{5^2} \right) \]
Calculating this gives:
\[ E = -13.6 \, \text{eV} \left( 1 - \frac{1}{25} \right) = -13.6 \, \text{eV} \left( \frac{24}{25} \right) = -13.6 \times 0.96 = -13.056 \, \text{eV} \]

### Step 3: Determine the second highest energy photon
The transitions from \(n=5\) to lower states will produce multiple photons. The second highest energy photon will be from the transition \(n=5\) to \(n=3\):
\[ E_{5 \to 3} = -13.6 \, \text{eV} \left( \frac{1}{3^2} - \frac{1}{5^2} \right) \]
Calculating this gives:
\[ E_{5 \to 3} = -13.6 \, \text{eV} \left( \frac{1}{9} - \frac{1}{25} \right) = -13.6 \, \text{eV} \left( \frac{25 - 9}{225} \right) = -13.6 \, \text{eV} \left( \frac{16}{225} \right) \]
Calculating this gives:
\[ E_{5 \to 3} = -13.6 \times 0.0711 = -0.968 \, \text{eV} \]

### Step 4: Use the photon energy to excite the Li^(2+) ion
The photon with energy \(E_{5 \to 3}\) can be used to excite an electron in the Li^(2+) ion. The energy levels of the Li^(2+) ion are given by:
\[ E = -\frac{Z^2 \cdot 13.6}{n^2} \, \text{eV} \]
For Li^(2+), \(Z = 3\):
\[ E = -\frac{3^2 \cdot 13.6}{n^2} = -\frac{122.4}{n^2} \]

### Step 5: Find the final excited state of Li^(2+)
We need to find \(n\) such that:
\[ -0.968 = -\frac{122.4}{n^2} \]
Rearranging gives:
\[ n^2 = \frac{122.4}{0.968} \approx 126.5 \]
Taking the square root:
\[ n \approx 11.25 \]
Since \(n\) must be an integer, we round to the nearest integer, which is \(n = 12\).

### Final Answer:
The final excited state of the Li^(2+) ion is \(n = 12\).

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