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. Here’s the step-by-step solution:
### Step 1: Understand the photoelectric effect
According to the photoelectric effect, when light falls on a metal surface, it can eject electrons. The energy of the incoming photons is given by \( E = h \nu \), where \( h \) is Planck's constant and \( \nu \) is the frequency of the light. The maximum kinetic energy (K.E.) of the emitted electrons can be calculated using the equation:
\[
K.E. = E - \phi
\]
where \( \phi \) is the work function of the material.
### Step 2: Calculate the maximum kinetic energy of the emitted electrons
Given:
- Photon energy \( E = 5.50 \, \text{eV} \)
- Work function \( \phi = 4.50 \, \text{eV} \)
Using the equation for kinetic energy:
\[
K.E. = 5.50 \, \text{eV} - 4.50 \, \text{eV} = 1.00 \, \text{eV}
\]
### Step 3: Relate kinetic energy to wavelength
The kinetic energy of the emitted electron can also be expressed in terms of its wavelength using the de Broglie wavelength formula:
\[
K.E. = \frac{h c}{\lambda}
\]
where \( c \) is the speed of light and \( \lambda \) is the wavelength. Rearranging this gives:
\[
\lambda = \frac{h c}{K.E.}
\]
### Step 4: Substitute the values
We need to convert the kinetic energy from eV to joules for calculation:
1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)
Thus,
\[
K.E. = 1.00 \, \text{eV} = 1.00 \times 1.6 \times 10^{-19} \, \text{J} = 1.6 \times 10^{-19} \, \text{J}
\]
Now, substituting the values:
- Planck's constant \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Speed of light \( c = 3.0 \times 10^{8} \, \text{m/s} \)
Now calculate \( \lambda \):
\[
\lambda = \frac{(6.63 \times 10^{-34} \, \text{Js}) \times (3.0 \times 10^{8} \, \text{m/s})}{1.6 \times 10^{-19} \, \text{J}}
\]
### Step 5: Calculate the wavelength
Calculating the numerator:
\[
6.63 \times 10^{-34} \times 3.0 \times 10^{8} = 1.989 \times 10^{-25} \, \text{Js m}
\]
Now dividing by the kinetic energy:
\[
\lambda = \frac{1.989 \times 10^{-25}}{1.6 \times 10^{-19}} = 1.243125 \times 10^{-6} \, \text{m}
\]
### Step 6: Convert to Angstroms
To convert meters to Angstroms (1 m = \( 10^{10} \) Angstroms):
\[
\lambda = 1.243125 \times 10^{-6} \, \text{m} \times 10^{10} \, \text{Å/m} = 124.3125 \, \text{Å}
\]
### Step 7: Round the answer
Rounding this value gives approximately:
\[
\lambda \approx 1240 \, \text{Å}
\]
### Final Answer
The wavelength of the fastest electron emitted is approximately **1240 Å**.
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