To find the stopping potential for the given setup, we can use the photoelectric equation:
\[ E = eV_0 = \frac{hc}{\lambda} - \Phi \]
Where:
- \( E \) is the energy of the incident photons,
- \( e \) is the charge of an electron,
- \( V_0 \) is the stopping potential,
- \( h \) 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 incident radiation,
- \( \Phi \) is the work function of the metal.
Given:
- Wavelength \( \lambda = 2000 \, \text{Å} = 2000 \times 10^{-10} \, \text{m} = 2 \times 10^{-7} \, \text{m} \)
- Work function \( \Phi = 5.01 \, \text{eV} \)
### Step 1: Calculate the energy of the incident photons \( E \)
Using the formula for energy:
\[ E = \frac{hc}{\lambda} \]
Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{2 \times 10^{-7} \, \text{m}} \]
### Step 2: Calculate \( E \)
Calculating the numerator:
\[ E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{2 \times 10^{-7}} \]
\[ E = \frac{1.9878 \times 10^{-25}}{2 \times 10^{-7}} \]
\[ E = 9.939 \times 10^{-19} \, \text{J} \]
### Step 3: Convert energy from Joules to electron volts
To convert Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{9.939 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \]
\[ E \approx 6.21 \, \text{eV} \]
### Step 4: Use the photoelectric equation to find the stopping potential \( V_0 \)
Now substituting back into the photoelectric equation:
\[ eV_0 = E - \Phi \]
\[ V_0 = \frac{E - \Phi}{e} \]
Substituting the values:
\[ V_0 = 6.21 \, \text{eV} - 5.01 \, \text{eV} \]
\[ V_0 = 1.20 \, \text{eV} \]
### Step 5: Conclusion
Thus, the stopping potential \( V_0 \) is approximately \( 1.20 \, \text{V} \).
### Final Answer:
The stopping potential for the given setup is approximately \( 1.20 \, \text{V} \).
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