To solve the problem, we will follow these steps:
### Step 1: Identify the parameters from the electric field equation
The electric field associated with the light wave is given by:
\[ E = E_0 \sin\left[(1.57 \times 10^7 \, \text{m}^{-1})(ct - x)\right] \]
From this equation, we can identify the angular frequency \( \omega \) and the wave number \( k \).
### Step 2: Relate wave number to wavelength
The wave number \( k \) is given by:
\[ k = 1.57 \times 10^7 \, \text{m}^{-1} \]
The wavelength \( \lambda \) can be calculated using the relation:
\[ \lambda = \frac{2\pi}{k} \]
Substituting the value of \( k \):
\[ \lambda = \frac{2\pi}{1.57 \times 10^7} \]
### Step 3: Calculate the wavelength
Calculating the wavelength:
\[ \lambda = \frac{2 \times 3.14}{1.57 \times 10^7} \approx \frac{6.28}{1.57 \times 10^7} \approx 4 \times 10^{-7} \, \text{m} \]
Converting this to nanometers:
\[ \lambda \approx 400 \, \text{nm} \]
### Step 4: Use the photoelectric effect equation
The maximum kinetic energy \( K.E. \) of the emitted electrons can be expressed using the photoelectric effect equation:
\[ K.E. = \frac{hc}{\lambda} - \phi \]
where \( \phi \) is the work function of the emitter. Given \( \phi = 2.1 \, \text{eV} \) and using \( h = 6.62 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \), we first calculate \( \frac{hc}{\lambda} \).
### Step 5: Calculate \( \frac{hc}{\lambda} \)
Substituting the values:
\[ \frac{hc}{\lambda} = \frac{(6.62 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{4 \times 10^{-7} \, \text{m}} \]
Calculating this:
\[ \frac{hc}{\lambda} = \frac{1.986 \times 10^{-25}}{4 \times 10^{-7}} \approx 4.965 \times 10^{-19} \, \text{J} \]
Now converting Joules to electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ \frac{hc}{\lambda} \approx \frac{4.965 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.103 \, \text{eV} \]
### Step 6: Calculate the maximum kinetic energy
Now, substituting into the kinetic energy equation:
\[ K.E. = 3.103 \, \text{eV} - 2.1 \, \text{eV} = 1.003 \, \text{eV} \]
### Step 7: Find the stopping potential
The stopping potential \( V_0 \) is equal to the maximum kinetic energy in electron volts:
\[ V_0 = K.E. = 1.003 \, \text{V} \]
### Final Answer
The stopping potential when this light is used in the photoelectric effect experiment is approximately:
\[ V_0 \approx 1.0 \, \text{V} \]
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