The work function of a metal is \(4.0\text{ eV}\). If the metal is irradiated with radiation of wavelength \(200\text{ nm}\), the maximum kinetic energy of the photoelectrons would be about: (Use \(hc = 1240\text{ eV}\cdot\text{nm}\))
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Using \(E = \frac{1240}{\lambda \text{ (in nm)}} \text{ eV}\) or \(E = \frac{12400}{\lambda \text{ (in \AA)}} \text{ eV}\) is a massive time-saver for calculating photon energy in Modern Physics problems.
Step 1: Understand what is being asked. Light of wavelength $200\text{ nm}$ falls on a metal whose work function is $4.0\text{ eV}$. We want the largest kinetic energy an ejected electron can carry away. Step 2: Write down the photoelectric idea. Each photon hands its whole energy to one electron. Part of that energy is spent freeing the electron (the work function $\Phi$), and whatever is left becomes kinetic energy: \[ K_{\max} = E_{\text{photon}} - \Phi \] Step 3: Find the photon energy. Photon energy is $E = \dfrac{hc}{\lambda}$. The problem kindly gives $hc = 1240\text{ eV}\cdot\text{nm}$, which keeps the arithmetic clean. Step 4: Plug in the wavelength. \[ E = \frac{1240\text{ eV}\cdot\text{nm}}{200\text{ nm}} = 6.2\text{ eV} \] So every incoming photon delivers $6.2\text{ eV}$. Step 5: Subtract the work function. \[ K_{\max} = 6.2\text{ eV} - 4.0\text{ eV} = 2.2\text{ eV} \] Step 6: State the answer. The fastest electrons leave with about $2.2\text{ eV}$ of kinetic energy, which is option (C). \[ \boxed{K_{\max} = 2.2\text{ eV}} \]