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Step-by-Step Solution
Step 1: Identify the frequency from the given electric field expression
The electric field of the light wave is given by
$$
\overrightarrow{E} = 10^{-3} \cos\left(\frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14}\, t\right)\
\hat{x}\,\frac{N}{C}.
$$
From this, note that the angular frequency $ \omega $ is
$$
\omega = 2\pi \times 6 \times 10^{14}.
$$
Therefore, the frequency $f$ is
$$
f = 6 \times 10^{14}\,\text{Hz}.
$$
Step 2: Calculate the wavelength of the incident light
We know the relationship between frequency $f$, wavelength $\lambda$, and the speed of light $c$:
$$
c = f \lambda.
$$
Here, $c = 3 \times 10^{8}\,\text{m/s}$ and $f = 6 \times 10^{14}\,\text{Hz}$. Hence,
$$
\lambda = \frac{c}{f}
= \frac{3 \times 10^{8}}{6 \times 10^{14}}
= 5 \times 10^{-7} \,\text{m}.
$$
Converting to Angström ($1\,\mathrm{Å} = 10^{-10}\,\mathrm{m}$):
$$
\lambda = 5000\,\mathrm{Å}.
$$
Step 3: Determine the photon energy
The energy of a photon (in electron-volts) can be calculated using the relation
$$
E = \frac{12375}{\lambda (\mathrm{\AA})}.
$$
Substituting $\lambda = 5000\,\mathrm{Å}$:
$$
E = \frac{12375}{5000} = 2.475\,\text{eV}.
$$
Step 4: Apply Einstein’s photoelectric equation to find the stopping potential
The maximum kinetic energy $K_{\text{max}}$ of the emitted photoelectrons is given by:
$$
K_{\text{max}} = E - \phi,
$$
where $E$ is the photon energy and $\phi$ is the work function (2 eV). Thus,
$$
K_{\text{max}} = 2.475\,\text{eV} - 2\,\text{eV} = 0.475\,\text{eV}.
$$
Step 5: Relate maximum kinetic energy to stopping potential
In the photoelectric effect, the stopping potential $V_{\text{s}}$ is the potential needed to stop the most energetic photoelectrons. Since the maximum kinetic energy in electron-volts is numerically equal to the stopping potential in volts,
$$
eV_{\text{s}} = K_{\text{max}} \quad \Longrightarrow \quad V_{\text{s}} = 0.475\,\text{V}.
$$
Rounding suitably, we get
$$
V_{\text{s}} \approx 0.48\,\text{V}.
$$
Final Answer: 0.48 V
