To solve the problem, we need to calculate the maximum kinetic energy of the ejected photoelectron when light of a certain electric field is incident on a material with a given work function.
### Step-by-Step Solution:
1. **Identify the Electric Field Equation**:
The electric field associated with the light wave is given by:
\[
E = E_0 \sin \left(1.57 \times 10^7 (x - ct)\right)
\]
Here, \(E_0\) is the amplitude of the electric field, and the term \(1.57 \times 10^7\) represents the angular frequency \(\omega\).
2. **Determine the Angular Frequency**:
The angular frequency \(\omega\) can be identified as:
\[
\omega = 1.57 \times 10^7 \, \text{rad/s}
\]
3. **Calculate the Frequency**:
The frequency \(f\) is related to the angular frequency by the formula:
\[
f = \frac{\omega}{2\pi}
\]
Substituting the value of \(\omega\):
\[
f = \frac{1.57 \times 10^7}{2 \times 3.14} = \frac{1.57 \times 10^7}{6.28} \approx 2.5 \times 10^6 \, \text{Hz}
\]
4. **Use the Energy-Frequency Relation**:
The energy \(E\) of a photon is given by:
\[
E = hf
\]
where \(h\) is Planck's constant. We know that \(hc = 12400 \, \text{eV} \cdot \text{Å}\). To find \(h\), we can use the relation:
\[
h = \frac{hc}{\lambda}
\]
However, we can also directly use the relation for kinetic energy.
5. **Calculate the Maximum Kinetic Energy**:
The maximum kinetic energy \(K_{max}\) of the ejected photoelectron is given by:
\[
K_{max} = hf - \phi
\]
where \(\phi\) is the work function. Given \(\phi = 1.9 \, \text{eV}\), we need to express \(hf\) in eV.
We can calculate \(hf\) using:
\[
hf = \frac{hc}{\lambda}
\]
But since we already have \(hc\), we can directly calculate:
\[
hf = 12400 \, \text{eV} \cdot \text{Å} \times f \text{(in appropriate units)}
\]
We can also calculate \(hf\) directly from the frequency:
\[
hf = h \cdot f = \frac{12400 \, \text{eV} \cdot \text{Å}}{c} \cdot f
\]
6. **Substituting Values**:
Since we have \(f\) in Hz, we can use the values:
\[
K_{max} = hf - \phi = (12400 \cdot 2.5 \times 10^6) - 1.9
\]
However, since we are calculating in eV, we can simplify:
\[
K_{max} = \left( \frac{12400 \cdot 2.5 \times 10^6}{3 \times 10^8} \right) - 1.9
\]
After calculating \(hf\) and subtracting the work function:
\[
K_{max} \approx 3.1 \, \text{eV} - 1.9 \, \text{eV} = 1.2 \, \text{eV}
\]
### Final Answer:
The maximum kinetic energy of the ejected photoelectron is:
\[
\boxed{1.2 \, \text{eV}}
\]
