Certainly! Let's find the work function of the metal given the longest wavelength of light that can cause photoelectron emission. The work function is the minimum energy required to emit an electron from a metal surface.
Given:
- Wavelength ([tex]\(\lambda\)[/tex]) = [tex]\(6.5 \times 10^{-7}\)[/tex] meters
To solve this, we will make use of the following constants:
- Planck's constant ([tex]\(h\)[/tex]) = [tex]\(6.626 \times 10^{-34}\)[/tex] J·s
- Speed of light ([tex]\(c\)[/tex]) = [tex]\(3 \times 10^8\)[/tex] m/s
- Conversion factor for Joules to electron volts (eV): [tex]\(1 \text{eV} = 1.602 \times 10^{-19}\)[/tex] J
### Step-by-Step Solution:
#### Step 1: Calculating the Energy in Joules
To find the energy [tex]\(E\)[/tex] of a photon corresponding to the given wavelength, we use the equation:
[tex]\[ E = \frac{hc}{\lambda} \][/tex]
Substitute the given values:
[tex]\[ h = 6.626 \times 10^{-34} \ \text{J·s} \][/tex]
[tex]\[ c = 3 \times 10^8 \ \text{m/s} \][/tex]
[tex]\[ \lambda = 6.5 \times 10^{-7} \ \text{m} \][/tex]
[tex]\[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{6.5 \times 10^{-7}} \][/tex]
Performing the division and multiplication, the energy [tex]\(E\)[/tex] is:
[tex]\[ E \approx 3.058 \times 10^{-19} \ \text{Joules} \][/tex]
So, the work function of the metal in joules is:
[tex]\[ \Phi = 3.058 \times 10^{-19} \ \text{J} \][/tex]
#### Step 2: Converting Energy from Joules to Electron Volts (eV)
Next, we convert this energy into electron volts. Using the conversion factor:
[tex]\[ 1 \ \text{eV} = 1.602 \times 10^{-19} \ \text{J} \][/tex]
The energy in electron volts (eV) is given by:
[tex]\[ \text{Energy in eV} = \frac{\text{Energy in J}}{\text{Conversion factor}} \][/tex]
So, convert the energy:
[tex]\[ \Phi \ (\text{in eV}) = \frac{3.058 \times 10^{-19}}{1.602 \times 10^{-19}} \][/tex]
Performing the division:
[tex]\[ \Phi \approx 1.909 \ \text{eV} \][/tex]
#### Summary:
The work function of the metal is:
- In Joules: [tex]\( 3.058 \ \times 10^{-19} \ \text{J} \)[/tex]
- In electron volts: [tex]\( 1.909 \ \text{eV} \)[/tex]
I hope this detailed step-by-step solution helps you understand the process of determining the work function!
