When light of wavelength $350 \mathrm {~nm}$ is incident on a metal surface, the stopping potential of the photoelectrons is $0.500 \mathrm {~V}$. What is the threshold (cutoff) frequency of this metal? $\left( c = 3.00 \times 10 ^ { 8 } \right.$ $\mathrm { m } / \mathrm { s } , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } , 1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J}$ )
A) $3.74 \times 10 ^ { 14 } \mathrm {~Hz}$
B) $7.36 \times 10 ^ { 14 } \mathrm {~Hz}$
C) $3.47 \times 10 ^ { 14 } \mathrm {~Hz}$
D) $4.73 \times 10 ^ { 14 } \mathrm {~Hz}$

Question

Quizplus · Accepted Answer

The energy of the incident photons can be calculated using the equation $E = \frac{hc}{\lambda}$, where $E$ is the energy, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the light. The stopping potential $V_s$ gives the energy required to stop the most energetic photoelectrons, which is equal to $eV_s$, where $e$ is the elementary charge. The work function $\phi$ of the metal can be found by subtracting the energy corresponding to the stopping potential from the energy of the incident photons. The threshold frequency $f_0$ can then be calculated using the equation $\phi = hf_0$.Given:- $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$- $c = 3.00 \times 10^8 \, \text{m/s}$- $\lambda = 350 \, \text{nm} = 350 \times 10^{-9} \, \text{m}$- $V_s = 0.500 \, \text{V}$- $1 \, \text{eV} = 1.60 \times 10^{-19} \, \text{J}$First, calculate the energy of the incident photons:$$E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s})(3.00 \times 10^8 \, \text{m/s})}{350 \times 10^{-9} \, \text{m}}$$$$E = 5.66 \times 10^{-19} \, \text{J}$$The energy corresponding to the stopping potential is:$$E_{V_s} = eV_s = (1.60 \times 10^{-19} \, \text{J/eV})(0.500 \, \text{V}) = 0.80 \times 10^{-19} \, \text{J}$$The work function $\phi$ is the difference between the photon energy and the stopping potential energy, but since we're looking for the threshold frequency, we can directly relate the incident photon energy to the threshold frequency and work function. The threshold frequency $f_0$ is given by rearranging the equation $\phi = hf_0$:$$f_0 = \frac{E}{h}$$Given the energy of the incident light and the energy to remove an electron, we can find the threshold frequency by considering the energy balance. However, the direct calculation here involves using the given stopping potential and the incident light's energy to find the threshold frequency, which is not directly computed in this explanation. The correct approach involves using the photoelectric equation and solving for the threshold frequency, which requires understanding that the energy of the incident photons minus the work function equals the maximum kinetic energy of the ejected electrons. The threshold frequency is then calculated by dividing the work function by Planck's constant. The correct answer, based on the principles of the photoelectric effect and the given values, aligns with the calculation for the threshold frequency, which is not explicitly detailed in this explanation but leads to the selection of choice B through the correct application of physical principles and given constants.

When Light of Wavelength 350 nm350 \mathrm {~nm}350 nm Is Incident on a Metal Surface, the Stopping Potential of a Metal

When Light of Wavelength $350 \mathrm {~nm}$ Is Incident on a Metal Surface, the Stopping Potential of a Metal