A light wave has an associated electric field given by E = E_0 \!...

The given electric field is $E = E_0 \sin[(1.57 \times 10^7 \text{ m}^{-1})(x - ct)]$.

Comparing this with the standard wave equation $E = E_0 \sin[k(x - ct)]$, the wave number is $k = 1.57 \times 10^7 \text{ m}^{-1}$.

Since $k = \dfrac{2\pi}{\lambda}$, the wavelength $\lambda$ is:

$\lambda = \dfrac{2\pi}{1.57 \times 10^7} = \dfrac{2 \times 3.14}{1.57 \times 10^7} = 4 \times 10^{-7} \text{ m} = 4000 \text{ \AA}$

The energy of the incident photon is:

$E = \dfrac{hc}{\lambda} = \dfrac{12400 \text{ eV \AA}}{4000 \text{ \AA}} = 3.1 \text{ eV}$

From Einstein's photoelectric equation, the maximum kinetic energy K_\max is:

K_\max = E - \Phi

Substituting the given work function $\Phi = 1.9 \text{ eV}$:

K_\max = 3.1 \text{ eV} - 1.9 \text{ eV} = 1.2 \text{ eV}

The stopping potential $V_0$ is given by eV_0 = K_\max, which yields:

$V_0 = 1.2 \text{ V}$