Light of wavelength 439 nm is incident on a metallic surface with a resultant photoelectric stopping potential of 1.2 V.What is the work function of the metal? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m) A)1.63 eV B)2.83 eV C)4.03 eV D)0.82 eV E)2.02 eV

Question

Quizplus · Accepted Answer

The work function ($\phi$) of the metal can be calculated using the photoelectric equation: $\phi = E - eV$, where $E$ is the energy of the incident photon, $eV$ is the energy corresponding to the stopping potential, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the incident light. First, calculate the energy of the incident photon: $E = \frac{hc}{\lambda}$. Substituting the given values, $E = \frac{6.63 \times 10^{-34} \text{ J⋅s} \times 3.00 \times 10^8 \text{ m/s}}{439 \times 10^{-9} \text{ m}}$. This gives $E \approx 4.53 \times 10^{-19} \text{ J}$. Converting this energy to electron volts (eV) by dividing by the charge of an electron (1.60 \times 10^{-19} J/eV) gives $E \approx 2.83 \text{ eV}$. The energy corresponding to the stopping potential is $eV = 1.2 \text{ eV}$ (since 1 eV = 1.60 \times 10^{-19} J and the stopping potential is 1.2 V). Therefore, the work function is $\phi = E - eV = 2.83 \text{ eV} - 1.2 \text{ eV} = 1.63 \text{ eV}$.

Light of Wavelength 439 Nm Is Incident on a Metallic