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Step-by-step Solution
Step 1: Identify the given data
• Stopping potential $V_0 = 0.5\,\text{V}$ (equivalent to $0.5\,\text{eV}$ of kinetic energy).
• Wavelength of incident photon $\lambda = 250 \,\text{nm} = 2500\,\text{\AA}$.
Step 2: Convert the given wavelength to photon energy
The energy $E$ of a photon is given by
$$
E = \frac{hc}{\lambda}
$$
Using the approximate relation $hc = 12400\,\text{eV}\cdot\text{\AA}$ (when $\lambda$ is in \AA):
$$
E = \frac{12400\,\text{eV}\cdot\text{\AA}}{2500\,\text{\AA}} = 4.96\,\text{eV}
$$
Step 3: Relate the kinetic energy to the work function
In the photoelectric effect, the maximum kinetic energy $K.E_{\max}$ of the emitted electron is
$$
K.E_{\max} = E - \omega_0
$$
where $E$ is the energy of the incident photon and $\omega_0$ is the work function of the metal. Here,
$$
K.E_{\max} = 0.5\,\text{eV}
$$
and
$$
E = 4.96\,\text{eV}
$$
Step 4: Calculate the work function
Substitute the values:
$$
0.5\,\text{eV} = 4.96\,\text{eV} - \omega_0
$$
Therefore,
$$
\omega_0 = 4.96\,\text{eV} - 0.5\,\text{eV} = 4.46\,\text{eV}
$$
Rounding this value appropriately gives about $4.5\,\text{eV}$.
Step 5: State the final answer
The work function of the metal is $4.5\,\text{eV}$.
