To solve the problem, we need to find the work function (φ) of the material from which the emitter is made, using the given wavelength of light and the stopping potential of the photoelectrons. Here’s a step-by-step solution:
### Step 1: Understand the relationship between the stopping potential, work function, and wavelength
The photoelectric effect can be described by the equation:
\[ eV_0 = \frac{hc}{\lambda} - \phi \]
Where:
- \( e \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) C),
- \( V_0 \) is the stopping potential (0.38 V),
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \) J·s),
- \( c \) is the speed of light (\( 3 \times 10^8 \) m/s),
- \( \lambda \) is the wavelength of the light (488 nm or \( 488 \times 10^{-9} \) m),
- \( \phi \) is the work function of the material.
### Step 2: Convert the stopping potential to energy in electron volts
The stopping potential \( V_0 \) is given as 0.38 V. Since 1 V = 1 J/C, we can directly use this value in our calculations.
### Step 3: Calculate the energy of the incident photons
Using the equation for the energy of a photon:
\[ E = \frac{hc}{\lambda} \]
Substituting the values:
- \( h = 6.63 \times 10^{-34} \, \text{J·s} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- \( \lambda = 488 \times 10^{-9} \, \text{m} \)
Calculating:
\[
E = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{488 \times 10^{-9}}
\]
\[
E \approx \frac{1.989 \times 10^{-25}}{488 \times 10^{-9}}
\]
\[
E \approx 4.07 \times 10^{-19} \, \text{J}
\]
### Step 4: Convert the energy from Joules to electron volts
To convert Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[
E \approx \frac{4.07 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.54 \, \text{eV}
\]
### Step 5: Substitute back into the photoelectric equation to find the work function
Now we can rearrange the equation to find the work function \( \phi \):
\[
\phi = E - eV_0
\]
Substituting the values:
\[
\phi = 2.54 \, \text{eV} - 0.38 \, \text{V}
\]
\[
\phi \approx 2.16 \, \text{eV}
\]
### Conclusion
The work function of the material from which the emitter is made is approximately \( 2.16 \, \text{eV} \).
