Light of wavelength `500 nm` is incident on a metal with work function `2.28 eV`. The de Broglie wavelength of the emitted electron is

Options

A. `le 2.8 xx 10^(-12) m`
B. `lt 2.8 xx 10^(-10) m`
C. `lt 2.8 xx 10^(-9) m`
D. `ge 2.8 xx 10^(-9) m`

Explanation

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:
- \( h \) 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 in meters (500 nm = \( 500 \times 10^{-9} \, \text{m} \))

Calculating the energy:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{500 \times 10^{-9} \, \text{m}} = \frac{1.9878 \times 10^{-25}}{500 \times 10^{-9}} = 3.9756 \times 10^{-19} \, \text{J} \]
To convert this energy from Joules to electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ E = \frac{3.9756 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.485 \, \text{eV} \]

### Step 2: Calculate the maximum kinetic energy of the emitted electron
The maximum kinetic energy (\( K_{max} \)) of the emitted electron can be calculated using the equation:
\[ K_{max} = E - \phi \]
where \( \phi \) is the work function of the metal (2.28 eV). Thus:
\[ K_{max} = 2.485 \, \text{eV} - 2.28 \, \text{eV} = 0.205 \, \text{eV} \]

### Step 3: Calculate the momentum of the emitted electron
The kinetic energy is related to momentum (\( p \)) by the equation:
\[ K = \frac{p^2}{2m} \]
Rearranging gives:
\[ p = \sqrt{2mK} \]
where \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \, \text{kg} \)) and \( K \) is the kinetic energy in Joules:
\[ K = 0.205 \, \text{eV} = 0.205 \times 1.6 \times 10^{-19} \, \text{J} \approx 3.28 \times 10^{-20} \, \text{J} \]
Now substituting into the momentum formula:
\[ p = \sqrt{2 \times (9.11 \times 10^{-31} \, \text{kg}) \times (3.28 \times 10^{-20} \, \text{J})} \approx \sqrt{5.96 \times 10^{-50}} \approx 7.73 \times 10^{-25} \, \text{kg m/s} \]

### Step 4: Calculate the de Broglie wavelength
The de Broglie wavelength (\( \lambda_{dB} \)) is given by:
\[ \lambda_{dB} = \frac{h}{p} \]
Substituting the values:
\[ \lambda_{dB} = \frac{6.626 \times 10^{-34} \, \text{Js}}{7.73 \times 10^{-25} \, \text{kg m/s}} \approx 8.58 \times 10^{-10} \, \text{m} = 0.858 \, \text{nm} \]

### Final Answer
The de Broglie wavelength of the emitted electron is approximately \( 0.858 \, \text{nm} \).

---