The hydrogen atom in its ground state is excited by means of monochromatic radiation. Its resulting spectrum has six different lines. These radiations are incident on a metal plate. It is observed that only two of them are responsible for photoelectric effect. if the ratio of maximum kinetic energy of photoelectrons is the two cases is 5 then find the work function of the metal.

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript while providing detailed explanations for each step.

### Step 1: Understanding the Excitation of the Hydrogen Atom
The hydrogen atom in its ground state (n=1) is excited to higher energy levels. The number of spectral lines produced when an atom transitions between energy levels can be determined using the formula:

\[ \text{Number of lines} = \frac{N(N-1)}{2} \]

where \(N\) is the number of energy levels involved. Given that there are 6 lines, we can set up the equation:

\[ \frac{N(N-1)}{2} = 6 \]

### Step 2: Solving for N
Multiplying both sides by 2 gives:

\[ N(N-1) = 12 \]

Now, we can test integer values for \(N\):

- For \(N = 4\): \(4(4-1) = 12\) (This works)

Thus, \(N = 4\), meaning the hydrogen atom can be excited to the first four energy levels (n=1, 2, 3, 4).

### Step 3: Energy Levels of Hydrogen Atom
The energy levels of a hydrogen atom are given by the formula:

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

Calculating the energies for \(n = 1, 2, 3, 4\):

- \(E_1 = -13.6 \, \text{eV}\)
- \(E_2 = -3.4 \, \text{eV}\)
- \(E_3 = -1.51 \, \text{eV}\)
- \(E_4 = -0.85 \, \text{eV}\)

### Step 4: Finding Energy Differences
The maximum energy difference (which corresponds to the maximum kinetic energy of the photoelectrons) occurs for the transition from \(n=4\) to \(n=1\) and from \(n=3\) to \(n=1\):

1. **For \(n=4\) to \(n=1\)**:
\[ \Delta E_1 = E_4 - E_1 = -0.85 - (-13.6) = 12.75 \, \text{eV} \]

2. **For \(n=3\) to \(n=1\)**:
\[ \Delta E_2 = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \, \text{eV} \]

### Step 5: Applying the Photoelectric Effect
According to the problem, the ratio of the maximum kinetic energies of the photoelectrons emitted by the two wavelengths is given as:

\[ \frac{K_1}{K_2} = 5 \]

Using Einstein's photoelectric equation:

\[ K = E - \phi \]

where \(E\) is the energy of the incident photon and \(\phi\) is the work function of the metal, we can write:

\[ \frac{\Delta E_1 - \phi}{\Delta E_2 - \phi} = 5 \]

Substituting the values of \(\Delta E_1\) and \(\Delta E_2\):

\[ \frac{12.75 - \phi}{12.09 - \phi} = 5 \]

### Step 6: Solving for the Work Function \(\phi\)
Cross-multiplying gives:

\[ 12.75 - \phi = 5(12.09 - \phi) \]

Expanding and rearranging:

\[ 12.75 - \phi = 60.45 - 5\phi \]
\[ 5\phi - \phi = 60.45 - 12.75 \]
\[ 4\phi = 47.7 \]
\[ \phi = \frac{47.7}{4} = 11.925 \, \text{eV} \]

### Final Answer
The work function of the metal is:

\[ \phi = 11.925 \, \text{eV} \]