Co operative Supervisor & Environment Supervisor This question relates to the photoelectric effect. The relationship between photon energy ($E$), work function ($\phi$), and the kinetic energy of the emitted photoelectron ($K_{max}$) is given by Einstein's photoelectric equation: $E = \phi + K_{max}$. To emit a photoelectron with zero velocity, $K_{max} = 0$. Therefore, the energy of the incident photon must be equal to the work function: $E = \phi$. The work function is given as 4 eV. We need to convert this to Joules: $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$. So, $\phi = 4 \text{ eV} \times 10^{-19} \text{ J}$. The energy of a photon is also related to its wavelength ($\lambda$) by the equation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant ($6.626 \times 10^{-34} \text{ J} \cdot \text{s}$) and $c$ is the speed of light ($3 \times 10^8 \text{ m/s}$). Therefore, $\lambda = \frac{hc}{E} = \frac{hc}{\phi}$. $\lambda = \frac{(6.626 \times 10^{-34} \text{ J} \cdot \text{s}) \times (3 \times 10^8 \text{ m/s})}{6.408 \times 10^{-19} \text{ J}}$ $\lambda \approx 3.096 \times 10^{-7} \text{ m}$. Converting this to Angstroms ($\AA$), where $1 \text{ m} = 10^{10} \AA$: $\lambda \approx 3.096 \times 10^{-7} \text{ m} \times 10^{10} \AA/\text{m} \approx 3096 \AA$. The closest option to 3096 $\AA$ is 3108 $\AA$.