In a photo electric set up the radiation from Balmer series of hydrogen atoms are Incident on a metal of work function 2 eV. Find wavelength of incident radiations lying between 450 nm to 700 nm. Find the maximum kinetic energy of the photoelectron emitted (hc1242 eV nm).

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript while ensuring clarity in each step.

### Step 1: Understanding the Balmer Series
The Balmer series corresponds to the transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 3) to the second energy level (n = 2). The energy of the emitted photons can be calculated using the formula:

\[ E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For the Balmer series, \( n_1 = 2 \) and \( n_2 \) can be 3, 4, 5, etc.

### Step 2: Calculate the Energy for Different Transitions
We will calculate the energy for the first few transitions in the Balmer series:

1. **Transition from n=3 to n=2:**
\[ E_{3 \to 2} = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{9 - 4}{36} \right) = 13.6 \left( \frac{5}{36} \right) \approx 1.89 \, \text{eV} \]

2. **Transition from n=4 to n=2:**
\[ E_{4 \to 2} = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \left( \frac{4 - 1}{16} \right) = 13.6 \left( \frac{3}{16} \right) \approx 2.55 \, \text{eV} \]

3. **Transition from n=5 to n=2:**
\[ E_{5 \to 2} = 13.6 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{25} \right) = 13.6 \left( \frac{25 - 4}{100} \right) = 13.6 \left( \frac{21}{100} \right) \approx 2.856 \, \text{eV} \]

### Step 3: Determine the Energy Range
From the calculations:
- The energy for \( n=3 \to n=2 \) is approximately 1.89 eV.
- The energy for \( n=4 \to n=2 \) is approximately 2.55 eV.
- The energy for \( n=5 \to n=2 \) is approximately 2.856 eV.

Given the work function \( \phi = 2 \, \text{eV} \), we can see that the energies of 2.55 eV and 2.856 eV are sufficient to emit photoelectrons.

### Step 4: Calculate Maximum Kinetic Energy
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be calculated using the equation:

\[ \text{K.E.}_{\text{max}} = E - \phi \]

Using \( E = 2.55 \, \text{eV} \):

\[ \text{K.E.}_{\text{max}} = 2.55 \, \text{eV} - 2 \, \text{eV} = 0.55 \, \text{eV} \]

### Step 5: Calculate the Wavelength of the Incident Radiation
To find the wavelength corresponding to the energy of 2.55 eV, we can use the formula:

\[ E = \frac{hc}{\lambda} \]

Rearranging gives:

\[ \lambda = \frac{hc}{E} \]

Substituting \( h = 1242 \, \text{eV nm} \) and \( E = 2.55 \, \text{eV} \):

\[ \lambda = \frac{1242 \, \text{eV nm}}{2.55 \, \text{eV}} \approx 487.45 \, \text{nm} \]

### Final Answers
- **Wavelength of incident radiation:** \( \approx 487.45 \, \text{nm} \)
- **Maximum kinetic energy of the photoelectron:** \( 0.55 \, \text{eV} \)