The electric field of light wave is given as $$\vec{E} = 10^{-3} \cos\left(\frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14}t\right) \hat{x} \frac{N}{C}$$. This light falls on a metal plate of work function 2 eV. The stopping potential of the photo-electrons is: Given, E (in eV) = $$\frac{12375}{\lambda(\text{in } \mathring{A})}$$

We begin by reading the mathematical form of the light wave. The electric‐field vector is written as

$$\vec E \;=\; 10^{-3}\; \cos\!\left(\frac{2\pi x}{5\times10^{-7}} - 2\pi \times 6\times10^{14}\,t\right)\,\hat x\;\frac{\text N}{\text C}.$$

Inside the cosine we recognise the term $$\frac{2\pi x}{\lambda}$$, where $$\lambda$$ is the wavelength of the wave. Comparing, we have

$$\lambda \;=\; 5\times10^{-7}\,\text m.$$

To use the given energy-wavelength relation, we must convert the wavelength from metres to ångströms. Since

$$1\ \text{\AA} = 10^{-10}\,\text m,$$

we write

$$\lambda \;=\; \frac{5\times10^{-7}\,\text m}{10^{-10}\,\text m/\text{\AA}} = 5\times10^{-7 + 10}\,\text{\AA} = 5\times10^{3}\,\text{\AA} = 5000\,\text{\AA}.$$

The problem supplies the convenient formula

$$E_\gamma\ (\text{in eV}) = \frac{12375}{\lambda\ (\text{in \AA})}.$$

Substituting $$\lambda = 5000\ \text{\AA}$$, we obtain the photon energy:

$$E_\gamma = \frac{12375}{5000} = 2.475\ \text{eV}.$$

The metal plate has a work function $$\phi = 2.00\ \text{eV}.$$ According to Einstein’s photoelectric equation, the maximum kinetic energy $$K_{\text{max}}$$ of the emitted photo-electrons is the excess energy of the photon over the work function, that is,

$$K_{\text{max}} = E_\gamma - \phi = 2.475\ \text{eV} - 2.00\ \text{eV} = 0.475\ \text{eV}.$$

The stopping potential $$V_0$$ is defined by the relation

$$e\,V_0 = K_{\text{max}},$$

where $$e$$ is the elementary charge. Because $$K_{\text{max}}$$ is already expressed in electron-volts, dividing by $$e$$ simply converts the energy in eV to volts:

$$V_0 = \frac{K_{\text{max}}}{e} = 0.475\ \text{V}.$$

Rounding to two significant figures (in line with the data supplied), we write

$$V_0 \approx 0.48\ \text{V}.$$

Hence, the correct answer is Option D.