A sample of hydrogen atom is in excited state (all the atoms ). The photons emitted from this sample are made to pass through a filter through which light having wavelength greater than `800 nm` can only pass. Only one type of photons are found to pass the filter. The sample's excited state initially is `[Take h c = 1240 eV -nm `, ground state energy of hydrogen atom `= - 13.6 eV]`.

Options

A. Fifth excited state
B. Fourth excited state
C. third excited state
D. Second excited state

Explanation

To solve the problem step by step, we will analyze the information provided and calculate the necessary values.

### Step 1: Understand the problem
We have a sample of hydrogen atoms in an excited state, and we need to determine the initial excited state of the atoms based on the photons emitted and passing through a filter that allows only wavelengths greater than 800 nm.

### Step 2: Calculate the energy corresponding to 800 nm
Using the formula for energy of a photon:
\[ E = \frac{hc}{\lambda} \]
Given that \(hc = 1240 \, \text{eV} \cdot \text{nm}\) and \(\lambda = 800 \, \text{nm}\):
\[ E = \frac{1240 \, \text{eV} \cdot \text{nm}}{800 \, \text{nm}} = 1.55 \, \text{eV} \]

### Step 3: Determine the energy levels of hydrogen
The energy levels of the hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
Calculating the first few energy levels:
- For \(n = 1\): \(E_1 = -13.6 \, \text{eV}\)
- For \(n = 2\): \(E_2 = -3.4 \, \text{eV}\)
- For \(n = 3\): \(E_3 = -1.51 \, \text{eV}\)
- For \(n = 4\): \(E_4 = -0.85 \, \text{eV}\)
- For \(n = 5\): \(E_5 = -0.54 \, \text{eV}\)

### Step 4: Identify the transition that produces photons with energy less than 1.55 eV
Since the filter allows only photons with wavelengths greater than 800 nm, the corresponding energy must be less than 1.55 eV. Thus, we need to find transitions from higher energy levels to lower ones that yield energies less than 1.55 eV.

### Step 5: Check possible transitions
- Transition from \(n = 4\) to \(n = 3\):
\[ E_{4 \to 3} = E_3 - E_4 = (-1.51) - (-0.85) = -1.51 + 0.85 = -0.66 \, \text{eV} \quad (\text{valid}) \]
- Transition from \(n = 3\) to \(n = 2\):
\[ E_{3 \to 2} = E_2 - E_3 = (-3.4) - (-1.51) = -3.4 + 1.51 = -1.89 \, \text{eV} \quad (\text{valid}) \]
- Transition from \(n = 2\) to \(n = 1\):
\[ E_{2 \to 1} = E_1 - E_2 = (-13.6) - (-3.4) = -13.6 + 3.4 = -10.2 \, \text{eV} \quad (\text{valid}) \]

### Step 6: Determine the initial excited state
The only transition that corresponds to a photon energy less than 1.55 eV and can pass through the filter is from \(n = 4\) to \(n = 3\). Therefore, the initial excited state of the hydrogen atom must be \(n = 4\).

### Conclusion
The sample's initial excited state is \(n = 4\).