If the work function of a metal surface is $2.20 \mathrm { eV }$, what frequency of incident light would give a maximum kinetic energy of $0.25 \mathrm { eV }$ to the photoelectrons ejected from this surface? $( 1 \mathrm { eV } = 1.60 \times$ $\left. 10 ^{- 19} \mathrm {~J} , h = 6.626 \times 10 ^{- 34}\mathrm {~J} \cdot \mathrm { s } \right)$ A) $1.02 \times 10 ^ { 14 } \mathrm {~Hz}$ B) $2.05 \times 10 ^ { 14 } \mathrm {~Hz}$ C) $5.92 \times 10 ^ { 14 } \mathrm {~Hz}$ D) $2.50 \times 10 ^ { 14 } \mathrm {~Hz}$ E) $3.53 \times 10 ^ { 14 } \mathrm {~Hz}$

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Quizplus · Accepted Answer

The energy of the incident light ($E$) is the sum of the work function ($\phi$) and the maximum kinetic energy ($K_{\text{max}}$) of the photoelectrons. Thus, $E = \phi + K_{\text{max}}$. Given $\phi = 2.20 \, \text{eV}$ and $K_{\text{max}} = 0.25 \, \text{eV}$, $E = 2.20 \, \text{eV} + 0.25 \, \text{eV} = 2.45 \, \text{eV}$. Converting $E$ to joules ($1 \, \text{eV} = 1.60 \times 10^{-19} \, \text{J}$), $E = 2.45 \, \text{eV} \times 1.60 \times 10^{-19} \, \text{J/eV} = 3.92 \times 10^{-19} \, \text{J}$. Using the equation $E = h\nu$, where $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$, solve for $\nu$: $\nu = \frac{E}{h} = \frac{3.92 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}} \approx 5.92 \times 10^{14} \, \text{Hz}$.

If the Work Function of a Metal Surface Is 2.20eV2.20 \mathrm { eV }2.20eV

If the Work Function of a Metal Surface Is $2.20 \mathrm { eV }$