To solve the problem step by step, we need to find the wavelength of incident light required to emit a photoelectron from a metal surface with a work function of 4 eV, where the emitted photoelectron has zero velocity.
### Step 1: Understand the Work Function
The work function (W₀) is the minimum energy required to remove an electron from the surface of the metal. Given:
\[ W₀ = 4 \, \text{eV} \]
### Step 2: Relate Energy to Wavelength
The energy of a photon can be expressed in terms of its wavelength (λ) using the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.
### Step 3: Set Up the Energy Equation
For a photoelectron to be emitted with zero velocity, all the energy of the photon must be used to overcome the work function:
\[ E = W₀ \]
Thus, we can write:
\[ \frac{hc}{\lambda} = W₀ \]
### Step 4: Convert Work Function to Joules
Since the work function is given in electron volts, we need to convert it to joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
So,
\[ W₀ = 4 \, \text{eV} = 4 \times 1.6 \times 10^{-19} \, \text{J} = 6.4 \times 10^{-19} \, \text{J} \]
### Step 5: Rearrange the Equation to Solve for Wavelength
Now, rearranging the equation for wavelength (λ):
\[ \lambda = \frac{hc}{W₀} \]
### Step 6: Substitute the Values
Substituting the known values:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- \( W₀ = 6.4 \times 10^{-19} \, \text{J} \)
So,
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{6.4 \times 10^{-19} \, \text{J}} \]
### Step 7: Calculate the Wavelength
Calculating the above expression:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{6.4 \times 10^{-19}} \]
\[ \lambda \approx 3.1 \times 10^{-7} \, \text{m} \]
Converting to angstroms (1 m = \( 10^{10} \) angstroms):
\[ \lambda \approx 3100 \, \text{angstroms} \]
### Final Answer
The wavelength of the incident light should be approximately:
\[ \lambda \approx 3100 \, \text{angstroms} \]
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