The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface , having work function `5.01 eV` , when ultraviolet light of `200 nm` falls on it , must be

Options

A. 2.4 V
B. – 1.2 V
C. ) – 2.4 V
D. 1.2 V

Explanation

To solve the problem of finding the potential difference required to stop the fastest photoelectrons emitted by a nickel surface when ultraviolet light of 200 nm falls on it, we can follow these steps:

### Step 1: Calculate the energy of the incident photons
The energy of a photon can be calculated using the formula:
\[ E = \frac{1240}{\lambda} \]
where \(E\) is the energy in electron volts (eV) and \(\lambda\) is the wavelength in nanometers (nm).

Given:
\(\lambda = 200 \, \text{nm}\)

Substituting the value:
\[ E = \frac{1240}{200} = 6.2 \, \text{eV} \]

### Step 2: Determine the work function of nickel
The work function (\(\Phi\)) of nickel is given as:
\[ \Phi = 5.01 \, \text{eV} \]

### Step 3: Calculate the maximum kinetic energy of the emitted photoelectrons
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be calculated using the equation:
\[ \text{K.E.} = E - \Phi \]

Substituting the values we found:
\[ \text{K.E.} = 6.2 \, \text{eV} - 5.01 \, \text{eV} = 1.19 \, \text{eV} \]

### Step 4: Relate kinetic energy to stopping potential
The stopping potential (\(V_0\)) required to stop the fastest photoelectrons can be expressed as:
\[ eV_0 = \text{K.E.} \]
where \(e\) is the charge of the electron.

Since we are looking for \(V_0\):
\[ V_0 = \text{K.E.} \]

Thus:
\[ V_0 = 1.19 \, \text{V} \]

### Conclusion
The potential difference that must be applied to stop the fastest photoelectrons emitted by the nickel surface is approximately:
\[ \boxed{1.19 \, \text{V}} \]