To solve the problem of finding 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, we can follow these steps:
### Step 1: Calculate the energy of the incident photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the light (in meters).
First, convert the wavelength from nanometers to meters:
\[ \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \]
Now, 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:
\[ E \approx 3.976 \times 10^{-19} \, \text{J} \]
### Step 2: Convert the energy from Joules to electronvolts
To convert energy from Joules to electronvolts, use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus:
\[ E \text{ (in eV)} = \frac{3.976 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 2.485 \, \text{eV} \]
### Step 3: Determine the kinetic energy of the emitted electron
The kinetic energy (KE) of the emitted electron can be found using the equation:
\[ KE = E_{\text{photon}} - \phi \]
where:
- \( \phi \) is the work function of the metal (2.28 eV).
Substituting the values:
\[ KE = 2.485 \, \text{eV} - 2.28 \, \text{eV} \]
\[ KE \approx 0.205 \, \text{eV} \]
### Step 4: Calculate the de Broglie wavelength of the emitted electron
The de Broglie wavelength \( \lambda_d \) of a particle can be calculated using the formula:
\[ \lambda_d = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum can be expressed in terms of kinetic energy:
\[ p = \sqrt{2m_e KE} \]
where \( m_e \) is the mass of the electron (\( 9.11 \times 10^{-31} \, \text{kg} \)).
First, convert the kinetic energy from eV to Joules:
\[ KE = 0.205 \, \text{eV} = 0.205 \times 1.6 \times 10^{-19} \, \text{J} \approx 3.28 \times 10^{-20} \, \text{J} \]
Now, calculate the momentum:
\[ p = \sqrt{2 \times 9.11 \times 10^{-31} \, \text{kg} \times 3.28 \times 10^{-20} \, \text{J}} \]
Calculating this gives:
\[ p \approx \sqrt{5.96 \times 10^{-50}} \approx 7.73 \times 10^{-25} \, \text{kg m/s} \]
Finally, substitute \( p \) back into the de Broglie wavelength formula:
\[ \lambda_d = \frac{6.626 \times 10^{-34} \, \text{Js}}{7.73 \times 10^{-25} \, \text{kg m/s}} \]
Calculating this gives:
\[ \lambda_d \approx 8.58 \times 10^{-10} \, \text{m} = 0.858 \, \text{nm} \]
### Final Answer
The de Broglie wavelength of the emitted electron is approximately **0.858 nm**.
