Light of wavelength 400 nm bombards a surface that has a work function of 2.46 eV. What is the maximum kinetic energy of the photoelectrons from this surface?
A) No photoelectrons are emitted.
B) 2.46 eV
C) 3.30 eV
D) 0.640 eV
E) 0.840 eV

Question

Quizplus · Accepted Answer

The maximum kinetic energy of the photoelectrons can be calculated using the photoelectric equation: $KE_{max} = hf - \phi$, where $KE_{max}$ is the maximum kinetic energy of the photoelectrons, $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{m}^2 \, \text{kg} / \text{s}$), $f$ is the frequency of the incident light, and $\phi$ is the work function of the material. The frequency $f$ can be found from the wavelength $\lambda$ using the relation $f = \frac{c}{\lambda}$, where $c$ is the speed of light ($3.00 \times 10^8 \, \text{m/s}$). Given $\lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}$, and $\phi = 2.46 \, \text{eV}$, we first convert the work function to joules ($1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$), then calculate $KE_{max}$.First, convert the wavelength to frequency:$$f = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \, \text{m/s}}{400 \times 10^{-9} \, \text{m}} = 7.5 \times 10^{14} \, \text{Hz}$$Then, calculate the energy of the photons using $E = hf$:$$E = (6.626 \times 10^{-34} \, \text{J s})(7.5 \times 10^{14} \, \text{Hz}) = 4.9695 \times 10^{-19} \, \text{J}$$Convert the work function to joules:$$\phi = 2.46 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 3.94092 \times 10^{-19} \, \text{J}$$Finally, calculate $KE_{max}$:$$KE_{max} = E - \phi = 4.9695 \times 10^{-19} \, \text{J} - 3.94092 \times 10^{-19} \, \text{J} = 1.02858 \times 10^{-19} \, \text{J}$$To convert $KE_{max}$ to electronvolts:$$KE_{max} = \frac{1.02858 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 0.640 \, \text{eV}$$

Light of Wavelength 400 Nm Bombards a Surface That Has