Photon having energy equivalent to the binding energy of `4th` state of `He^(+)` atom is used to eject an electron from the metal surface of work function `1.4 eV`. If electrons are further accelerated through the potential difference of `4V` then the minimum value of De-brogile wavelength associated with the electron is:

Explanation

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

### Step 1: Calculate the Binding Energy of the 4th State of He⁺
The binding energy (BE) for an electron in a hydrogen-like atom is given by the formula:
\[ BE = \frac{13.6 Z^2}{n^2} \text{ eV} \]
For He⁺, \( Z = 2 \) and \( n = 4 \):
\[ BE = \frac{13.6 \times 2^2}{4^2} = \frac{13.6 \times 4}{16} = \frac{54.4}{16} = 3.4 \text{ eV} \]

### Step 2: Determine the Kinetic Energy of the Ejected Electron
The energy of the photon used to eject the electron is equal to the binding energy of the 4th state of He⁺, which we calculated as 3.4 eV. The work function (ϕ) of the metal is given as 1.4 eV. The kinetic energy (KE) of the ejected electron can be calculated using:
\[ KE = E_{photon} - \phi \]
Substituting the values:
\[ KE = 3.4 \text{ eV} - 1.4 \text{ eV} = 2.0 \text{ eV} \]

### Step 3: Calculate the Additional Kinetic Energy from Acceleration
The electron is accelerated through a potential difference of 4 V, which adds to its kinetic energy. The additional kinetic energy gained (KE_additional) is given by:
\[ KE_{additional} = e \cdot V = 1 \text{ eV} \cdot 4 \text{ V} = 4 \text{ eV} \]
Thus, the total kinetic energy (KE_total) of the electron after acceleration is:
\[ KE_{total} = KE + KE_{additional} = 2 \text{ eV} + 4 \text{ eV} = 6 \text{ eV} \]

### Step 4: Convert Kinetic Energy to Joules
To use the kinetic energy in the de Broglie wavelength formula, we need to convert the energy from electron volts to joules. The conversion factor is:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
Thus,
\[ KE_{total} = 6 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 9.6 \times 10^{-19} \text{ J} \]

### Step 5: Calculate the De Broglie Wavelength
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2 m KE}} \]
Where:
- \( h = 6.63 \times 10^{-34} \text{ J s} \) (Planck's constant)
- \( m = 9.1 \times 10^{-31} \text{ kg} \) (mass of the electron)

Substituting the values:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 9.6 \times 10^{-19}}} \]

Calculating the denominator:
\[ 2 \times 9.1 \times 10^{-31} \times 9.6 \times 10^{-19} = 1.74672 \times 10^{-48} \]
\[ \sqrt{1.74672 \times 10^{-48}} \approx 1.32 \times 10^{-24} \]

Now substituting back into the wavelength formula:
\[ \lambda = \frac{6.63 \times 10^{-34}}{1.32 \times 10^{-24}} \approx 5.01 \times 10^{-10} \text{ m} = 5.01 \text{ Å} \]

### Final Answer
The minimum value of the de Broglie wavelength associated with the electron is approximately:
\[ \lambda \approx 5 \text{ Å} \]