To find the maximum velocity of the electron emitted from the lithium surface when ultraviolet light of wavelength 280 nm is incident on it, we can use the principles of the photoelectric effect. Here is the step-by-step solution:
### Step 1: Calculate the frequency of the incident light
The frequency (\( \nu \)) can be calculated using the formula:
\[
\nu = \frac{c}{\lambda}
\]
where:
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \))
- \( \lambda \) is the wavelength in meters (\( 280 \, \text{nm} = 280 \times 10^{-9} \, \text{m} \))
Substituting the values:
\[
\nu = \frac{3 \times 10^8}{280 \times 10^{-9}} \approx 1.071 \times 10^{15} \, \text{Hz}
\]
### Step 2: Calculate the energy of the incident photons
The energy (\( E \)) of the photons can be calculated using the formula:
\[
E = h \nu
\]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \))
Substituting the values:
\[
E = 6.626 \times 10^{-34} \times 1.071 \times 10^{15} \approx 7.1 \times 10^{-19} \, \text{J}
\]
### Step 3: Convert the energy from Joules to electron volts
To convert energy from Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[
E \text{ (in eV)} = \frac{7.1 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 4.44 \, \text{eV}
\]
### Step 4: Determine the kinetic energy of the emitted electrons
The kinetic energy (\( KE \)) of the emitted electrons can be determined using the equation:
\[
KE = E - \phi
\]
where \( \phi \) is the work function of lithium (given as \( 2.5 \, \text{eV} \)):
\[
KE = 4.44 \, \text{eV} - 2.5 \, \text{eV} = 1.94 \, \text{eV}
\]
### Step 5: Convert kinetic energy to Joules
Convert the kinetic energy from electron volts to Joules:
\[
KE \text{ (in J)} = 1.94 \times 1.6 \times 10^{-19} \approx 3.104 \times 10^{-19} \, \text{J}
\]
### Step 6: Calculate the maximum velocity of the emitted electrons
Using the kinetic energy formula:
\[
KE = \frac{1}{2} mv^2
\]
we can solve for \( v \):
\[
v = \sqrt{\frac{2 \times KE}{m}}
\]
where \( m \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \)):
\[
v = \sqrt{\frac{2 \times 3.104 \times 10^{-19}}{9.1 \times 10^{-31}}}
\]
Calculating this gives:
\[
v \approx \sqrt{6.82 \times 10^{11}} \approx 8.25 \times 10^5 \, \text{m/s}
\]
### Final Answer:
The maximum velocity of the electron emitted from the lithium surface is approximately \( 8.25 \times 10^5 \, \text{m/s} \).
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