The energy of emitted electrons, known as the kinetic energy (KE), is given by the formula derived from the photoelectric effect:

$KE = hf - \phi$

where:

• $KE$ is the kinetic energy of the emitted electrons (7 eV in this case),
• $h$ is Planck’s constant ($4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s}$),
• $f$ is the frequency of the light (given as $2.5 \times 10^{15} \, \text{Hz}$),
• $\phi$ is the work function of the metal (the value we need to find).

Rearranging the formula to solve for $\phi$:

$\phi = hf - KE$

Step 1: Calculate $hf$

$hf = (4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s}) \times (2.5 \times 10^{15} \, \text{Hz})$ $hf = 10.33916924 \, \text{eV}$

Step 2: Substitute $hf$ and $KE$ into the equation for $\phi$

$\phi = 10.33916924 \, \text{eV} - 7 \, \text{eV}$ $\phi = 3.33916924 \, \text{eV}$

Final Answer

The work function $\phi$ of the metal is approximately:

$\phi \approx 3.34 \, \text{eV}$

Would you like further details or have any questions?

Here are some related questions for further exploration:

1. How does the work function relate to the threshold frequency of light for a metal?
2. What is the significance of Planck’s constant in the photoelectric effect?
3. How would the energy of emitted electrons change if the frequency of light were increased?
4. Can a metal emit electrons if the light frequency is below the threshold frequency? Why or why not?
5. How is the photoelectric effect applied in modern technology, such as in solar panels?

Tip: Always keep track of your units when working with physical constants and equations to ensure consistency in your calculations.