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 outlined in the video transcript while providing a clear structure for the calculations.

### Step 1: Understand the Problem
The problem states that a hydrogen atom in its ground state is excited, resulting in six spectral lines. Two of these lines are responsible for the photoelectric effect, and the ratio of the maximum kinetic energies of the photoelectrons from these two lines is given as 5.

### Step 2: Set Up the Kinetic Energy Equation
Let the maximum kinetic energies of the photoelectrons from the two transitions be \( K_1 \) and \( K_2 \). According to the problem, we have:
\[ \frac{K_1}{K_2} = 5 \]
This implies:
\[ K_1 = 5 K_2 \]

### Step 3: Relate Kinetic Energy to Energy Transitions
The kinetic energy of photoelectrons can be expressed in terms of the energy of the incident photons and the work function \( W \) of the metal:
\[ K_1 = E_1 - W \quad \text{and} \quad K_2 = E_2 - W \]
where \( E_1 \) and \( E_2 \) are the energies of the photons corresponding to the two transitions.

### Step 4: Substitute the Kinetic Energy Expressions
Substituting the expressions for \( K_1 \) and \( K_2 \) into the ratio, we get:
\[ \frac{E_1 - W}{E_2 - W} = 5 \]

### Step 5: Cross Multiply
Cross-multiplying gives:
\[ E_1 - W = 5(E_2 - W) \]
Expanding this, we have:
\[ E_1 - W = 5E_2 - 5W \]
Rearranging terms leads to:
\[ E_1 = 5E_2 - 4W \]

### Step 6: Determine Energy Levels
From the hydrogen atom, the energy levels are given by:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For \( n = 4 \):
\[ E_4 = -\frac{13.6}{16} = -0.85 \, \text{eV} \]
For \( n = 3 \):
\[ E_3 = -\frac{13.6}{9} = -1.51 \, \text{eV} \]
For \( n = 2 \):
\[ E_2 = -\frac{13.6}{4} = -3.4 \, \text{eV} \]
For \( n = 1 \):
\[ E_1 = -13.6 \, \text{eV} \]

### Step 7: Calculate the Energy Differences
Now we calculate the energy differences for the transitions:
1. From \( n = 4 \) to \( n = 1 \):
\[ \Delta E_1 = E_4 - E_1 = -0.85 - (-13.6) = 12.75 \, \text{eV} \]
2. From \( n = 3 \) to \( n = 1 \):
\[ \Delta E_2 = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \, \text{eV} \]

### Step 8: Substitute Energy Differences into the Equation
Substituting these values into our earlier equation:
\[ \frac{12.75 - W}{12.09 - W} = 5 \]

### Step 9: Cross Multiply Again
Cross-multiplying gives:
\[ 12.75 - W = 5(12.09 - W) \]
Expanding this:
\[ 12.75 - W = 60.45 - 5W \]
Rearranging gives:
\[ 4W = 60.45 - 12.75 \]
\[ 4W = 47.7 \]
\[ W = \frac{47.7}{4} = 11.925 \, \text{eV} \]

### Conclusion
The work function \( W \) of the metal is approximately \( 11.925 \, \text{eV} \).