In sample of excited hydrogen atoms, electron make transition from n = 2 to n = 1. Emitted quanta strike on a metal of work functio `[phi]4.2 ev`. Calculate the wavelength (in `Å`) associated with ejected electrons having maximum kinetic energy.

Answer

To solve the problem step by step, we will follow the outlined process:

### Step 1: Calculate the energy of the emitted photon (E)
The energy of the photon emitted when an electron transitions from n=2 to n=1 in a hydrogen atom can be calculated using the formula:

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

Here, \( n_{\text{final}} = 1 \) and \( n_{\text{initial}} = 2 \).

\[ E = 13.6 \, \text{eV} \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \, \text{eV} \left( 1 - \frac{1}{4} \right) = 13.6 \, \text{eV} \left( \frac{3}{4} \right) \]

Calculating this gives:

\[ E = 13.6 \times 0.75 = 10.2 \, \text{eV} \]

### Step 2: Relate the photon energy to work function and kinetic energy
The energy of the emitted photon is related to the work function (\( \phi \)) of the metal and the maximum kinetic energy (\( KE_{\text{max}} \)) of the ejected electrons by the equation:

\[ E = \phi + KE_{\text{max}} \]

Given that the work function \( \phi = 4.2 \, \text{eV} \), we can rearrange the equation to find \( KE_{\text{max}} \):

\[ KE_{\text{max}} = E - \phi = 10.2 \, \text{eV} - 4.2 \, \text{eV} = 6.0 \, \text{eV} \]

### Step 3: Calculate the wavelength associated with the ejected electrons
The wavelength (\( \lambda \)) of the ejected electrons can be calculated using the formula:

\[ \lambda = \frac{150}{KE_{\text{max}}} \]

Substituting \( KE_{\text{max}} = 6.0 \, \text{eV} \):

\[ \lambda = \frac{150}{6.0} = 25 \, \text{Å} \]

### Final Result
The wavelength associated with the ejected electrons having maximum kinetic energy is:

\[ \lambda = 25 \, \text{Å} \]