The photoelectrons are emitted when light of wavelength 150 nm falls on a metal surface of work function 5. 28 eV. What must be the potential difference (in V) applied to stop the fastest photoelectrons
(Take `h = 4.14 xx 10^(-15) eVs`)

Explanation

To solve the problem of finding the potential difference required to stop the fastest photoelectrons emitted from a metal surface when light of wavelength 150 nm falls on it, we will follow these steps:

### Step 1: Calculate the energy of the incident photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 4.14 \times 10^{-15} \, \text{eV s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda = 150 \, \text{nm} = 150 \times 10^{-9} \, \text{m} \)

Substituting the values:
\[ E = \frac{(4.14 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{150 \times 10^{-9} \, \text{m}} \]

Calculating this gives:
\[ E = \frac{(4.14 \times 3) \times 10^{-7}}{150} \, \text{eV} \]
\[ E = \frac{12.42 \times 10^{-7}}{150} \, \text{eV} = 8.28 \, \text{eV} \]

### Step 2: Apply the photoelectric equation
The photoelectric equation relates the energy of the incident photon to the work function of the metal and the maximum kinetic energy of the emitted photoelectrons:
\[ E = \phi + K_{\text{max}} \]
where:
- \( \phi = 5.28 \, \text{eV} \) (work function)
- \( K_{\text{max}} = eV_0 \) (maximum kinetic energy, where \( V_0 \) is the stopping potential)

Rearranging gives:
\[ K_{\text{max}} = E - \phi \]

### Step 3: Calculate the maximum kinetic energy
Substituting the values we have:
\[ K_{\text{max}} = 8.28 \, \text{eV} - 5.28 \, \text{eV} = 3.00 \, \text{eV} \]

### Step 4: Relate kinetic energy to stopping potential
The maximum kinetic energy is also given by:
\[ K_{\text{max}} = eV_0 \]
where \( e \) is the charge of an electron (in this case, it cancels out since we are working in eV).

Thus, we have:
\[ V_0 = K_{\text{max}} = 3.00 \, \text{V} \]

### Final Answer
The potential difference required to stop the fastest photoelectrons is:
\[ \boxed{3 \, \text{V}} \]