To solve the problem of finding the wavelength of the fastest electron emitted when light with a photon energy of 5.50 eV falls on a tungsten surface with a work function of 4.50 eV, we can follow these steps:
### Step 1: Understand the Photoelectric Effect
The photoelectric effect states that when light (photons) hits a metal surface, electrons can be emitted if the energy of the photons is greater than the work function of the metal. The work function (Φ) is the minimum energy required to remove an electron from the surface of the metal.
### Step 2: Calculate the Kinetic Energy of the Emitted Electron
The energy of the incident photon (E) is given as 5.50 eV, and the work function (Φ) of tungsten is 4.50 eV. The kinetic energy (KE) of the emitted electron can be calculated using the formula:
\[
KE = E - \Phi
\]
Substituting the values:
\[
KE = 5.50 \, \text{eV} - 4.50 \, \text{eV} = 1.00 \, \text{eV}
\]
### Step 3: Relate Kinetic Energy to Wavelength
The kinetic energy of the emitted electron can also be expressed in terms of its wavelength (λ) using the de Broglie wavelength formula:
\[
KE = \frac{hc}{\lambda}
\]
Where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\))
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\))
However, since we are working with electron volts (eV), we can use the conversion factor:
\[
1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}
\]
Thus, the kinetic energy in joules is:
\[
KE = 1.00 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.6 \times 10^{-19} \, \text{J}
\]
### Step 4: Solve for Wavelength
Rearranging the equation for wavelength gives:
\[
\lambda = \frac{hc}{KE}
\]
Substituting the known values:
\[
\lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{1.6 \times 10^{-19} \, \text{J}}
\]
Calculating this gives:
\[
\lambda = \frac{1.9878 \times 10^{-25} \, \text{J m}}{1.6 \times 10^{-19} \, \text{J}} \approx 1.24 \times 10^{-6} \, \text{m} = 1240 \, \text{nm}
\]
### Step 5: Convert Wavelength to Angstroms
To convert meters to angstroms (1 m = \(10^{10}\) Å):
\[
\lambda = 1240 \, \text{nm} = 12400 \, \text{Å}
\]
### Final Answer
The wavelength of the fastest electron emitted is approximately **12400 Å**.
---
