Electrons in hydrogen like atom `(Z= 3)` make transition from the fifth to the fourth orbit and from the third orbit. The resulting radiation are incident normally on a metal plate and eject photoelectrons. The stopping potential for the photoelectrons ejected by the shorter wavelength is 3.95 volts. Calculate the work function of the metal and the stopping potential for the photoelectron ejected by longer wavelength
(Rydberg constant` = 1.094 xx10^(7) m^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Energy for the Transition from n=5 to n=4

The energy of an electron in a hydrogen-like atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

For the transition from \( n=5 \) to \( n=4 \):
\[ E_5 = -\frac{13.6 \times 3^2}{5^2} = -\frac{13.6 \times 9}{25} = -4.896 \text{ eV} \]
\[ E_4 = -\frac{13.6 \times 3^2}{4^2} = -\frac{13.6 \times 9}{16} = -7.65 \text{ eV} \]
Now, calculate the energy difference:
\[ \Delta E_{5 \to 4} = E_4 - E_5 = -7.65 - (-4.896) = -7.65 + 4.896 = 2.754 \text{ eV} \]

### Step 2: Calculate the Energy for the Transition from n=4 to n=3

For the transition from \( n=4 \) to \( n=3 \):
\[ E_3 = -\frac{13.6 \times 3^2}{3^2} = -\frac{13.6 \times 9}{9} = -13.6 \text{ eV} \]
Now, calculate the energy difference:
\[ \Delta E_{4 \to 3} = E_3 - E_4 = -13.6 - (-7.65) = -13.6 + 7.65 = 5.95 \text{ eV} \]

### Step 3: Determine the Work Function of the Metal

The stopping potential \( V_0 \) for the photoelectrons ejected by the shorter wavelength (from the transition \( n=4 \to n=3 \)) is given as 3.95 volts. The energy of the photon corresponding to this transition is:
\[ E = \Delta E_{4 \to 3} = 5.95 \text{ eV} \]
Using the photoelectric equation:
\[ E = W + eV_0 \]
where \( W \) is the work function of the metal. Rearranging gives:
\[ W = E - eV_0 = 5.95 - 3.95 = 2.00 \text{ eV} \]

### Step 4: Calculate the Stopping Potential for the Longer Wavelength Transition

Now, we need to find the stopping potential for the transition from \( n=5 \to n=4 \):
\[ E = \Delta E_{5 \to 4} = 2.754 \text{ eV} \]
Using the photoelectric equation again:
\[ E = W + eV_0 \]
Substituting the known values:
\[ 2.754 = 2.00 + eV_0 \]
Solving for \( eV_0 \):
\[ eV_0 = 2.754 - 2.00 = 0.754 \text{ eV} \]
Thus, the stopping potential \( V_0 \) is:
\[ V_0 = 0.754 \text{ volts} \]

### Final Answers:
- Work function of the metal \( W = 2.00 \text{ eV} \)
- Stopping potential for the longer wavelength \( V_0 = 0.754 \text{ volts} \)