What is the cutoff (threshold) frequency for a metal surface that has a work function of $5.42 \mathrm { eV }$ ? (1 $\left. \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J} , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } \right)$
A) $1.31 \times 10 ^ { 15 } \mathrm {~Hz}$
B) $5.02 \times 10^{15} \mathrm {~Hz}$
C) $3.01 \times 10^{15} \mathrm {~Hz}$
D) $6.04 \times 10^{15} \mathrm {~Hz}$
E) $2.01 \times 10^{15} \mathrm {~Hz}$

Question

Quizplus · Accepted Answer

The cutoff frequency ($f_c$) can be calculated using the formula $f_c = \frac{W}{h}$, where $W$ is the work function in joules and $h$ is Planck's constant. First, convert the work function from eV to joules: $5.42 \, \text{eV} = 5.42 \times 1.60 \times 10^{-19} \, \text{J} = 8.672 \times 10^{-19} \, \text{J}$. Then, calculate the cutoff frequency: $f_c = \frac{8.672 \times 10^{-19}}{6.626 \times 10^{-34}} = 1.308 \times 10^{15} \, \text{Hz}$.

What Is the Cutoff (Threshold) Frequency for a Metal Surface 5.42eV5.42 \mathrm { eV }5.42eV

What Is the Cutoff (Threshold) Frequency for a Metal Surface $5.42 \mathrm { eV }$