To solve the problem, we need to find the wavelength of the fastest electron emitted when light with a photon energy of 5.50 eV falls on the tungsten surface, which has a work function of 4.50 eV.
### Step-by-Step Solution:
1. **Understand the Photoelectric Effect**:
The energy of the incoming photons can be used to overcome the work function of the material and provide kinetic energy to the emitted electrons. The relationship is given by:
\[
K.E. = E_{\text{photon}} - \phi
\]
where \( K.E. \) is the kinetic energy of the emitted electrons, \( E_{\text{photon}} \) is the energy of the incoming photon, and \( \phi \) is the work function.
2. **Substitute the Known Values**:
Given:
- \( E_{\text{photon}} = 5.50 \, \text{eV} \)
- \( \phi = 4.50 \, \text{eV} \)
Now, substituting the values into the equation:
\[
K.E. = 5.50 \, \text{eV} - 4.50 \, \text{eV} = 1.00 \, \text{eV}
\]
3. **Convert Kinetic Energy to Joules**:
To find the wavelength, we need to convert the kinetic energy from electron volts to joules. The conversion factor is:
\[
1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}
\]
Therefore:
\[
K.E. = 1.00 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.6 \times 10^{-19} \, \text{J}
\]
4. **Use the Kinetic Energy to Find Wavelength**:
The kinetic energy of the emitted electrons can also be expressed in terms of wavelength using the de Broglie wavelength formula:
\[
K.E. = \frac{h c}{\lambda}
\]
Rearranging gives:
\[
\lambda = \frac{h c}{K.E.}
\]
5. **Substitute the Constants**:
Using the values:
- Planck's constant, \( h = 6.63 \times 10^{-34} \, \text{J s} \)
- Speed of light, \( c = 3.00 \times 10^{8} \, \text{m/s} \)
Substitute these into the equation:
\[
\lambda = \frac{(6.63 \times 10^{-34} \, \text{J s}) (3.00 \times 10^{8} \, \text{m/s})}{1.6 \times 10^{-19} \, \text{J}}
\]
6. **Calculate the Wavelength**:
\[
\lambda = \frac{1.989 \times 10^{-25}}{1.6 \times 10^{-19}} \approx 1.243125 \times 10^{-7} \, \text{m}
\]
Converting to angstroms (1 m = \( 10^{10} \) angstroms):
\[
\lambda \approx 12.43 \, \text{nm} = 1243 \, \text{Å}
\]
### Final Answer:
The wavelength of the fastest electron emitted is approximately **1243 Å**.
