Step 1: Understanding the Concept:
This problem involves the photoelectric effect, described by Einstein's photoelectric equation. When a photon strikes a metal surface, its energy is used to overcome the metal's work function (the minimum energy required to eject an electron), and any remaining energy is converted into the kinetic energy of the emitted electron (photoelectron).
Step 2: Key Formula or Approach:
1. Einstein's Photoelectric Equation: $K_{max} = E - \phi_0$, where:
- $K_{max}$ is the maximum kinetic energy of the photoelectron.
- $E$ is the energy of the incident photon.
- $\phi_0$ is the work function of the material.
2. Energy of a Photon: The energy of a photon is related to its wavelength ($\lambda$) by $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant and $c$ is the speed of light.
3. Useful Shortcut: When wavelength $\lambda$ is given in angstroms (\AA), the photon energy $E$ in electron-volts (eV) can be quickly calculated using the formula:
\[ E (\text{eV}) = \frac{12400}{\lambda (\text{\AA})} \]
Step 3: Detailed Explanation:
Given:
- Wavelength of light, $\lambda = 5000$ \AA.
- Work function, $\phi_0 = 1.9$ eV.
First, calculate the energy of the incident photons in eV using the shortcut formula:
\[ E = \frac{12400}{5000} = \frac{12.4}{5} = 2.48 \text{ eV} \]
Now, use Einstein's photoelectric equation to find the maximum kinetic energy of the emitted photoelectrons:
\[ K_{max} = E - \phi_0 \]
\[ K_{max} = 2.48 \text{ eV} - 1.9 \text{ eV} \]
\[ K_{max} = 0.58 \text{ eV} \]
Step 4: Final Answer:
The kinetic energy of the emitted photoelectron will be 0.58 eV. Therefore, option (A) is correct.