To find the maximum kinetic energy of the photoelectron, we need to use the photoelectric equation:
\(E_k = h\nu - \phi\),
where \(E_k\) is the kinetic energy of the emitted photoelectron, \(h\nu\) is the energy of the incident photon, and \(\phi\) is the work function of the metal.
- First, we convert the given wavelengths into energy using the formula: \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant and \(c\) is the speed of light.
- The threshold wavelength \(\lambda_0\) is \(4000\ \text{Å}\). The energy of photons with this wavelength (work function) is calculated as:
\(\phi = \frac{hc}{\lambda_0} = \frac{6.626 \times 10^{-34}\ \text{J}\cdot\text{s} \times 3 \times 10^8\ \text{m/s}}{4000 \times 10^{-10}\ \text{m}}\) - Calculate: \(\phi = 4.97 \times 10^{-19}\ \text{J}\) which is equal to \(3.1\ \text{eV}\) (since 1 eV = \(1.6 \times 10^{-19}\ \text{J}\)).
- The wavelength of the incident light \(\lambda\) is \(3000\ \text{Å}\). Its energy \(E\) is calculated as:
\(E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34}\ \text{J}\cdot\text{s} \times 3 \times 10^8\ \text{m/s}}{3000 \times 10^{-10}\ \text{m}}\) - Calculate: \(E = 6.63 \times 10^{-19}\ \text{J}\) which is approximately \(4.14\ \text{eV}\).
- Now substitute the values into the photoelectric equation:
\(E_k = 4.14\ \text{eV} - 3.1\ \text{eV}\) - Solve for \(E_k\):
\(E_k = 1.04\ \text{eV}\), which can be rounded off to \(1.0\ \text{eV}\) based on given options.
Therefore, the maximum kinetic energy of the photoelectron is \(1.0\ \text{eV}\). Thus, the correct answer is \(1.0\ \text{eV}\).