Hydrogen atom is its ground state is excited by means of monochromatic radiation of wavelength `1023 Å`. How many different lines are possible in the resulting spectrum? Calculate the longes wavelength among them. You may assume the ionization energy of hydrogen atom as 13.6 eV.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the energy of the incident radiation
The energy of the monochromatic radiation can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \, \text{eV s} \))
- \( c \) is the speed of light (\( 3 \times 10^{10} \, \text{cm/s} \) or \( 3 \times 10^{18} \, \text{Å/s} \))
- \( \lambda \) is the wavelength in Ångströms (given as \( 1023 \, \text{Å} \))

Substituting the values:
\[ E = \frac{(4.135667696 \times 10^{-15} \, \text{eV s}) \times (3 \times 10^{18} \, \text{Å/s})}{1023 \, \text{Å}} \]
Calculating this gives:
\[ E \approx 12.09 \, \text{eV} \]

### Step 2: Determine the possible energy levels the electron can reach
The ground state energy of the hydrogen atom is \( -13.6 \, \text{eV} \). The energy levels are given by:
\[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]
Where \( n \) is the principal quantum number.

Calculating the energy for the first few 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 \( n = 5 \): \( E_5 = -0.544 \, \text{eV} \)
- For \( n = 6 \): \( E_6 = -0.38 \, \text{eV} \)
- For \( n = 7 \): \( E_7 = -0.204 \, \text{eV} \)
- For \( n = 8 \): \( E_8 = -0.125 \, \text{eV} \)
- For \( n = 9 \): \( E_9 = -0.056 \, \text{eV} \)
- For \( n = 10 \): \( E_{10} = -0.04 \, \text{eV} \)

### Step 3: Determine the maximum principal quantum number reached
The energy absorbed by the electron is \( 12.09 \, \text{eV} \). The maximum energy level it can reach is when:
\[ E_n = E_1 + E = -13.6 + 12.09 = -1.51 \, \text{eV} \]
This corresponds to \( n = 3 \). Therefore, the electron can transition to levels \( n = 2 \) and \( n = 3 \).

### Step 4: Calculate the possible transitions
The possible transitions from the ground state (n=1) are:
1. \( 1 \rightarrow 2 \)
2. \( 1 \rightarrow 3 \)
3. \( 2 \rightarrow 3 \)

### Step 5: Calculate the wavelengths of the emitted lines
Using the formula for the wavelength:
\[ \lambda = \frac{hc}{\Delta E} \]
Where \( \Delta E \) is the energy difference between the levels.

1. For \( 1 \rightarrow 2 \):
\[ \Delta E = E_2 - E_1 = (-3.4) - (-13.6) = 10.2 \, \text{eV} \]
\[ \lambda_{1 \rightarrow 2} = \frac{12375}{10.2} \approx 1213 \, \text{Å} \]

2. For \( 1 \rightarrow 3 \):
\[ \Delta E = E_3 - E_1 = (-1.51) - (-13.6) = 12.09 \, \text{eV} \]
\[ \lambda_{1 \rightarrow 3} = \frac{12375}{12.09} \approx 1023 \, \text{Å} \]

3. For \( 2 \rightarrow 3 \):
\[ \Delta E = E_3 - E_2 = (-1.51) - (-3.4) = 1.89 \, \text{eV} \]
\[ \lambda_{2 \rightarrow 3} = \frac{12375}{1.89} \approx 6540 \, \text{Å} \]

### Step 6: Identify the longest wavelength
From the calculated wavelengths:
- \( 1213 \, \text{Å} \) (for \( 1 \rightarrow 2 \))
- \( 1023 \, \text{Å} \) (for \( 1 \rightarrow 3 \))
- \( 6540 \, \text{Å} \) (for \( 2 \rightarrow 3 \))

The longest wavelength is \( 6540 \, \text{Å} \).

### Final Answer
- The number of different lines possible in the resulting spectrum is **3**.
- The longest wavelength among them is **6540 Å**.