The electric field associated with a light wave is given by `E= E_0 sin [(1.57x 10^7 m^(-1)(x-ct)].` Find the stopping potential when this light is used in an experiment on photoelectric affect with a metal having work - function 1.9 eV.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given electric field equation
The electric field associated with the light wave is given by:
\[ E = E_0 \sin(1.57 \times 10^7 \, \text{m}^{-1} (x - ct)) \]

### Step 2: Determine the wave number \( K \)
From the equation, we can identify the wave number \( K \) as:
\[ K = 1.57 \times 10^7 \, \text{m}^{-1} \]

### Step 3: Calculate the wavelength \( \lambda \)
Using the relationship between wave number and wavelength:
\[ K = \frac{2\pi}{\lambda} \]
We can rearrange this to find \( \lambda \):
\[ \lambda = \frac{2\pi}{K} = \frac{2\pi}{1.57 \times 10^7} \]

Calculating \( \lambda \):
\[ \lambda \approx \frac{6.2832}{1.57 \times 10^7} \approx 4 \times 10^{-7} \, \text{m} \]

### Step 4: Calculate the energy of the incident photon
The energy \( E \) of a photon is given by:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h = 6.63 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \)

Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{4 \times 10^{-7}} \]

Calculating \( E \):
\[ E \approx \frac{1.989 \times 10^{-25}}{4 \times 10^{-7}} \approx 4.9725 \times 10^{-19} \, \text{J} \]

To convert this energy into electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ E \approx \frac{4.9725 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.11 \, \text{eV} \]

### Step 5: Determine the maximum kinetic energy \( K_{e \text{max}} \)
The maximum kinetic energy of the emitted electrons can be calculated using:
\[ K_{e \text{max}} = E - \phi \]
Where \( \phi \) is the work function of the metal (given as 1.9 eV).

Substituting the values:
\[ K_{e \text{max}} = 3.11 \, \text{eV} - 1.9 \, \text{eV} = 1.21 \, \text{eV} \]

### Step 6: Calculate the stopping potential \( V_0 \)
The stopping potential is given by:
\[ V_0 = K_{e \text{max}} \]
Thus,
\[ V_0 = 1.21 \, \text{V} \]

### Final Answer
The stopping potential when this light is used in an experiment on the photoelectric effect with a metal having a work function of 1.9 eV is approximately:
\[ V_0 \approx 1.21 \, \text{V} \]

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