Step 1: Identify the wave number from the given equation.
The electric field is given as $E = E_0 \sin[(1.57 \times 10^7\ \text{m}^{-1})(x - ct)]$. Comparing with the standard form $E = E_0 \sin(kx - \omega t)$, we identify the wave number as $k = 1.57 \times 10^7\ \text{m}^{-1}$.
Step 2: Find the wavelength of the light.
The wave number and wavelength are related by $k = \frac{2\pi}{\lambda}$. Therefore, \[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{1.57 \times 10^7} = \frac{6.28}{1.57 \times 10^7} = 4 \times 10^{-7}\ \text{m} = 400\ \text{nm} \] This is violet light in the visible spectrum.
Step 3: Calculate the energy of each photon.
Photon energy is $E = \frac{hc}{\lambda}$. Using $h = 6.64 \times 10^{-34}$ Js and $c = 3 \times 10^8$ m/s, \[ E = \frac{6.64 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} = \frac{19.92 \times 10^{-26}}{4 \times 10^{-7}} = 4.98 \times 10^{-19}\ \text{J} \]
Step 4: Convert photon energy to electron volts.
Since $1\ \text{eV} = 1.6 \times 10^{-19}$ J, \[ E = \frac{4.98 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.1\ \text{eV} \]
Step 5: Apply Einstein's photoelectric equation.
Einstein's photoelectric equation states that the maximum kinetic energy of the emitted photoelectron is \[ KE_{\text{max}} = E_{\text{photon}} - \phi \] where $\phi$ is the work function of the metal. Here $\phi = 1.9$ eV, so \[ KE_{\text{max}} = 3.1 - 1.9 = 1.2\ \text{eV} \approx 1.1\ \text{eV} \]
Step 6: Relate maximum kinetic energy to stopping potential.
The stopping potential $V_0$ is defined such that $eV_0 = KE_{\text{max}}$. Since the charge of an electron is $e = 1$ (in eV units), the stopping potential numerically equals the maximum KE in eV. \[ V_0 = 1.1\ \text{V} \] \[ \boxed{\text{Stopping potential} = 1.1\ \text{eV}} \]