To solve the problem, we need to find the de Broglie wavelength of the emitted electron when light of wavelength 500 nm is incident on a metal with a work function of 2.28 eV.
### Step-by-Step Solution:
**Step 1: Calculate the energy of the incident photon.**
The energy \( E \) of a photon can be calculated using the formula:
\[
E = \frac{hc}{\lambda}
\]
where:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
Substituting the values:
\[
E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{500 \times 10^{-9} \, \text{m}}
\]
Calculating this gives:
\[
E \approx 3.976 \times 10^{-19} \, \text{J}
\]
To convert this energy into electron volts (eV), we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[
E \approx \frac{3.976 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 2.485 \, \text{eV}
\]
**Step 2: Calculate the kinetic energy of the emitted electron.**
The kinetic energy \( KE \) of the emitted electron can be found using the equation:
\[
KE = E - \phi
\]
where \( \phi \) is the work function of the metal.
Given \( \phi = 2.28 \, \text{eV} \):
\[
KE = 2.485 \, \text{eV} - 2.28 \, \text{eV} \approx 0.205 \, \text{eV}
\]
**Step 3: Calculate the de Broglie wavelength of the emitted electron.**
The de Broglie wavelength \( \lambda_{dB} \) is given by:
\[
\lambda_{dB} = \frac{h}{p}
\]
where \( p \) is the momentum of the electron. The momentum can be expressed in terms of kinetic energy:
\[
p = \sqrt{2m \cdot KE}
\]
where \( m = 9.11 \times 10^{-31} \, \text{kg} \) (mass of the electron).
Substituting the values:
\[
p = \sqrt{2 \cdot (9.11 \times 10^{-31} \, \text{kg}) \cdot (0.205 \, \text{eV} \cdot 1.6 \times 10^{-19} \, \text{J/eV})}
\]
Calculating this gives:
\[
p \approx \sqrt{2 \cdot (9.11 \times 10^{-31}) \cdot (3.28 \times 10^{-20})} \approx \sqrt{5.96 \times 10^{-50}} \approx 7.73 \times 10^{-25} \, \text{kg m/s}
\]
Now substituting \( p \) into the de Broglie wavelength formula:
\[
\lambda_{dB} = \frac{6.626 \times 10^{-34} \, \text{Js}}{7.73 \times 10^{-25} \, \text{kg m/s}} \approx 8.58 \times 10^{-10} \, \text{m} = 0.858 \, \text{nm}
\]
### Final Answer:
The de Broglie wavelength of the emitted electron is approximately \( 0.858 \, \text{nm} \).
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