Hydrogen gas in the atomic state is excited to an energy level such tht the electrostatic potential energy of H-atom becomes `-1.7 eV`. Now, a photoelectric plate having work function w=2.3 eV is exposed to the emission spectra of this gas. Assuming all the transitions to be possible, find the minimum de-Broglie wavelength of the ejected photoelectrons.

Explanation

To solve the problem step-by-step, we will follow the reasoning outlined in the video transcript and derive the necessary values.

### Step 1: Understand the potential energy of the hydrogen atom
The problem states that the electrostatic potential energy (U) of the hydrogen atom in the excited state is given as:
\[ U = -1.7 \, \text{eV} \]

### Step 2: Relate potential energy to kinetic energy
In quantum mechanics, the kinetic energy (KE) of an electron in a hydrogen atom can be related to the potential energy. The relationship is given by:
\[ KE = -\frac{U}{2} \]
Substituting the value of U:
\[ KE = -\frac{-1.7}{2} = 0.85 \, \text{eV} \]

### Step 3: Use the formula for kinetic energy in terms of principal quantum number
The kinetic energy can also be expressed in terms of the principal quantum number \( n \):
\[ KE = -\frac{13.6}{n^2} \]
Setting these two expressions for kinetic energy equal gives:
\[ 0.85 = -\frac{13.6}{n^2} \]
Solving for \( n^2 \):
\[ n^2 = \frac{-13.6}{0.85} \]
Calculating \( n^2 \):
\[ n^2 \approx 16 \]
Thus, \( n \approx 4 \).

### Step 4: Calculate the energy difference for the transition
For the minimum de Broglie wavelength, we consider the transition from \( n = 4 \) to \( n = 1 \):
The energy levels for hydrogen are given by:
\[ E_n = -\frac{13.6}{n^2} \]
Calculating the energies:
- For \( n = 4 \):
\[ E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \, \text{eV} \]
- For \( n = 1 \):
\[ E_1 = -13.6 \, \text{eV} \]

The change in energy (\( \Delta E \)) is:
\[ \Delta E = E_4 - E_1 = -0.85 - (-13.6) = 12.75 \, \text{eV} \]

### Step 5: Calculate the maximum kinetic energy of the ejected photoelectrons
The maximum kinetic energy of the ejected photoelectrons can be calculated using the formula:
\[ KE_{max} = \Delta E - W \]
where \( W \) is the work function of the photoelectric plate:
\[ W = 2.3 \, \text{eV} \]
Thus,
\[ KE_{max} = 12.75 - 2.3 = 10.45 \, \text{eV} \]

### Step 6: Calculate the minimum de Broglie wavelength
The de Broglie wavelength (\( \lambda \)) of an electron can be calculated using the formula:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the electron, and can be related to kinetic energy:
\[ p = \sqrt{2m_e KE_{max}} \]
However, for simplicity, we can use the derived formula for the wavelength in terms of kinetic energy:
\[ \lambda = \sqrt{\frac{150}{KE_{max}}} \, \text{(in angstroms)} \]
Substituting \( KE_{max} = 10.45 \, \text{eV} \):
\[ \lambda = \sqrt{\frac{150}{10.45}} \]
Calculating \( \lambda \):
\[ \lambda \approx 3.8 \, \text{angstroms} \]

### Final Answer
The minimum de Broglie wavelength of the ejected photoelectrons is approximately:
\[ \lambda \approx 3.8 \, \text{angstroms} \]