Electron in a hydrogen-like atom `(Z = 3)` make transition from the forth excited state to the third excited state and from the third excited state to the second excited state. The resulting radiations are incident potential for photoelectrons ejested by shorter wavelength is `3.95 e V`.
Calculate the work function of the metal and stopping potiential for the photoelectrons ejected by the longer wavelength.

Options

A. `2.0 V`
B. `0.75 V`
C. `0.6 V`
D. none of these

Explanation

To solve the problem step by step, we need to calculate the energy transitions for the electron in a hydrogen-like atom with \( Z = 3 \) and then determine the work function and stopping potential for the photoelectrons.

### Step 1: Calculate the energy transition from the fourth excited state to the third excited state

The energy difference \( \Delta E_1 \) for the transition from the fourth excited state (\( n = 5 \)) to the third excited state (\( n = 4 \)) can be calculated using the formula:

\[ \Delta E_1 = 13.6 \, \text{eV} \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( Z = 3 \)
- \( n_1 = 4 \)
- \( n_2 = 5 \)

Substituting the values:

\[ \Delta E_1 = 13.6 \cdot 3^2 \left( \frac{1}{4^2} - \frac{1}{5^2} \right) \]

Calculating:

\[ \Delta E_1 = 13.6 \cdot 9 \left( \frac{1}{16} - \frac{1}{25} \right) \]

Finding a common denominator (which is 80):

\[ \Delta E_1 = 13.6 \cdot 9 \left( \frac{25 - 16}{400} \right) = 13.6 \cdot 9 \cdot \frac{9}{400} \]

\[ \Delta E_1 = \frac{13.6 \cdot 81}{400} = \frac{1099.6}{400} = 2.749 \, \text{eV} \approx 2.75 \, \text{eV} \]

### Step 2: Calculate the energy transition from the third excited state to the second excited state

Now, we calculate the energy difference \( \Delta E_2 \) for the transition from the third excited state (\( n = 4 \)) to the second excited state (\( n = 3 \)):

\[ \Delta E_2 = 13.6 \, \text{eV} \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( n_1 = 3 \)
- \( n_2 = 4 \)

Substituting the values:

\[ \Delta E_2 = 13.6 \cdot 3^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]

Calculating:

\[ \Delta E_2 = 13.6 \cdot 9 \left( \frac{1}{9} - \frac{1}{16} \right) \]

Finding a common denominator (which is 144):

\[ \Delta E_2 = 13.6 \cdot 9 \left( \frac{16 - 9}{144} \right) = 13.6 \cdot 9 \cdot \frac{7}{144} \]

\[ \Delta E_2 = \frac{13.6 \cdot 63}{144} \approx 5.95 \, \text{eV} \]

### Step 3: Determine the work function of the metal

We know that the energy of the incident radiation for the shorter wavelength corresponds to the transition from the third to the second excited state, which is given as \( 3.95 \, \text{eV} \):

Using the photoelectric equation:

\[ E_{\text{incident}} = \Delta E_2 = 5.95 \, \text{eV} \]

The work function \( \Phi \) can be calculated as:

\[ E_{\text{incident}} = \Phi + K.E. \]

Where \( K.E. = 3.95 \, \text{eV} \):

\[ 3.95 = 5.95 - \Phi \]

Rearranging gives:

\[ \Phi = 5.95 - 3.95 = 2.00 \, \text{eV} \]

### Step 4: Calculate the stopping potential for the longer wavelength

For the longer wavelength, the energy corresponds to the transition from the fourth excited state to the third excited state:

\[ E_{\text{incident}} = \Delta E_1 = 2.75 \, \text{eV} \]

Using the photoelectric equation again:

\[ E_{\text{incident}} = \Phi + K.E. \]

Let \( V_0 \) be the stopping potential:

\[ 2.75 = 2.00 + V_0 \]

Rearranging gives:

\[ V_0 = 2.75 - 2.00 = 0.75 \, \text{eV} \]

### Final Answers:
- Work function \( \Phi = 2.00 \, \text{eV} \)
- Stopping potential \( V_0 = 0.75 \, \text{eV} \)