A gas containining hydrogen-like ions, with atomic number `Z`, emits photons in transition `n + 2 to n, where n = Z` These photons fall on a metallic plate and eject electrons having minimum de Broglie wavelength `lambda of 5 Å`. Find the value of `Z` , if the work function of metal is `4.2 eV`

Explanation

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the Transition and Energy Calculation
The gas contains hydrogen-like ions with atomic number \( Z \). The transition of interest is from \( n + 2 \) to \( n \), where \( n = Z \). The energy of the photon emitted during this transition can be calculated using the formula:

\[ E = 13.6 \, Z^2 \left( \frac{1}{n^2} - \frac{1}{(n+2)^2} \right) \]

Substituting \( n = Z \):

\[ E = 13.6 \, Z^2 \left( \frac{1}{Z^2} - \frac{1}{(Z+2)^2} \right) \]

### Step 2: Simplify the Energy Expression
We can simplify the expression:

\[ E = 13.6 \, Z^2 \left( \frac{(Z+2)^2 - Z^2}{Z^2(Z+2)^2} \right) \]

Calculating \( (Z+2)^2 - Z^2 \):

\[ (Z+2)^2 - Z^2 = Z^2 + 4Z + 4 - Z^2 = 4Z + 4 \]

Thus, the energy becomes:

\[ E = 13.6 \, Z^2 \cdot \frac{4(Z + 1)}{Z^2(Z + 2)^2} = \frac{54.4(Z + 1)}{(Z + 2)^2} \]

### Step 3: Relate Energy to Work Function and Kinetic Energy
The emitted photons hit a metallic plate, ejecting electrons. The energy of the photons must equal the work function plus the kinetic energy of the ejected electrons:

\[ E = \text{Work Function} + K.E. \]

Given that the work function is \( 4.2 \, \text{eV} \) and the minimum de Broglie wavelength \( \lambda = 5 \, \text{Å} = 5 \times 10^{-10} \, \text{m} \), we can calculate the kinetic energy using the de Broglie wavelength formula:

\[ K.E. = \frac{h^2}{2m\lambda^2} \]

Where \( h \) (Planck's constant) is approximately \( 6.626 \times 10^{-34} \, \text{Js} \) and \( m \) (mass of an electron) is approximately \( 9.11 \times 10^{-31} \, \text{kg} \).

### Step 4: Calculate Kinetic Energy
Substituting the values:

\[ K.E. = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31} \times (5 \times 10^{-10})^2} \]

Calculating this gives:

\[ K.E. \approx 6 \, \text{eV} \]

### Step 5: Set Up the Equation
Now we can set up the equation:

\[ E = 4.2 + 6 = 10.2 \, \text{eV} \]

Thus, we equate:

\[ \frac{54.4(Z + 1)}{(Z + 2)^2} = 10.2 \]

### Step 6: Solve for Z
Cross-multiplying gives:

\[ 54.4(Z + 1) = 10.2(Z + 2)^2 \]

Expanding and rearranging leads to:

\[ 10.2(Z^2 + 4Z + 4) - 54.4Z - 54.4 = 0 \]

This simplifies to a quadratic equation:

\[ 10.2Z^2 - 4Z - 4 = 0 \]

### Step 7: Use the Quadratic Formula
Using the quadratic formula \( Z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Where \( a = 10.2, b = -4, c = -4 \):

Calculating the discriminant:

\[ D = (-4)^2 - 4 \cdot 10.2 \cdot (-4) = 16 + 163.2 = 179.2 \]

Now substituting into the formula:

\[ Z = \frac{4 \pm \sqrt{179.2}}{2 \cdot 10.2} \]

Calculating gives us two possible values for \( Z \), but only the positive integer solution is valid.

### Final Answer
After solving, we find:

\[ Z = 2 \]