What P.D. must be applied to stop the fastest photo electron emitted by a nickel surface when illuminated by ultraviolet light of wave length `2000^(@)A`. The work function of nickel is 5 eV

Options

A. 5 V
B. 6.2 V
C. 1.2 V
D. 1.2 V

Answer

To solve the problem of determining the potential difference (P.D.) required to stop the fastest photoelectron emitted by a nickel surface when illuminated by ultraviolet light of wavelength \(2000 \, \text{Å}\), we can follow these steps:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect states that when light of sufficient energy strikes a material, it can eject electrons from that material. The energy of the incoming photons must overcome the work function of the material to release electrons.

### Step 2: Calculate the Energy of the Incoming Photons
The energy \(E\) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h\) is Planck's constant (\(4.14 \times 10^{-15} \, \text{eV s}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

First, convert the wavelength from angstroms to meters:
\[ \lambda = 2000 \, \text{Å} = 2000 \times 10^{-10} \, \text{m} = 2 \times 10^{-7} \, \text{m} \]

Now, substitute the values into the energy formula:
\[ E = \frac{(4.14 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{2 \times 10^{-7} \, \text{m}} \]

### Step 3: Simplify the Calculation
Calculating the numerator:
\[ E = \frac{(4.14 \times 3) \times 10^{-7} \, \text{eV m}}{2 \times 10^{-7} \, \text{m}} = \frac{12.42 \times 10^{-7}}{2 \times 10^{-7}} \, \text{eV} = 6.21 \, \text{eV} \]

### Step 4: Determine the Kinetic Energy of the Ejected Electrons
The kinetic energy \(K.E.\) of the emitted photoelectrons is given by:
\[ K.E. = E - \text{Work Function} \]
Given that the work function of nickel is \(5 \, \text{eV}\):
\[ K.E. = 6.21 \, \text{eV} - 5 \, \text{eV} = 1.21 \, \text{eV} \]

### Step 5: Calculate the Stopping Potential
The stopping potential \(V_s\) required to stop the fastest photoelectron is equal to the kinetic energy of the emitted electrons in volts:
\[ V_s = K.E. = 1.21 \, \text{V} \]

### Final Answer
The potential difference that must be applied to stop the fastest photoelectron emitted by the nickel surface is approximately:
\[ \boxed{1.21 \, \text{V}} \]