A photocathode having a work function of $2.8 \mathrm { eV }$ is illuminated with monochromatic electromagnetic radiation whose photon energy is $4.0 \mathrm { eV }$. What is the threshold (cutoff) frequency for photoelectron production? $\left( 1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J} , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } \right)$
A) $8.6 \times 10 ^ { 14 } \mathrm {~Hz}$
B) $2.9 \times 10 ^ { 14 } \mathrm {~Hz}$
C) $6.8 \times 10 ^ { 14 } \mathrm {~Hz}$
D) $9.7 \times 10 ^ { 14 } \mathrm {~Hz}$
E) $7.7 \times 10 ^ { 14 } \mathrm {~Hz}$

Question

Quizplus · Accepted Answer

The threshold frequency is calculated using the formula $ f_0 = \frac{\phi}{h} $, where $\phi$ is the work function. Converting the work function from eV to Joules: $2.8 \, \text{eV} = 2.8 \times 1.60 \times 10^{-19} \, \text{J} = 4.48 \times 10^{-19} \, \text{J}$. Then, $ f_0 = \frac{4.48 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}} \approx 6.8 \times 10^{14} \, \text{Hz} $.

A Photocathode Having a Work Function Of 2.8eV2.8 \mathrm { eV }2.8eV

A Photocathode Having a Work Function Of $2.8 \mathrm { eV }$