To find the work function (Φ) of the metal in the photoelectric effect experiment, we can follow these steps:
### Step 1: Calculate the energy of the incident photon
The energy of the photon (E) can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) = Planck's constant = \( 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) = speed of light = \( 3 \times 10^8 \, \text{m/s} \)
- \( \lambda \) = wavelength of the radiation = \( 250 \, \text{nm} = 250 \times 10^{-9} \, \text{m} \)
### Step 2: Substitute the values into the formula
Substituting the values into the equation:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{250 \times 10^{-9} \, \text{m}} \]
### Step 3: Calculate the energy in joules
Calculating the above expression:
\[ E = \frac{(6.626 \times 3) \times 10^{-26}}{250} \]
\[ E = \frac{19.878 \times 10^{-26}}{250} \]
\[ E = 7.9512 \times 10^{-28} \, \text{J} \]
### Step 4: Convert energy from joules to electron volts
To convert joules to electron volts, we use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus, the energy in electron volts is:
\[ E \, (\text{eV}) = \frac{7.9512 \times 10^{-28} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \]
\[ E \approx 4.969 \, \text{eV} \]
### Step 5: Use the stopping potential to find the work function
The stopping potential (V) is given as 0.5 V, which corresponds to 0.5 eV. The relationship between the work function (Φ), photon energy (E), and stopping potential (V) is given by:
\[ E = \Phi + V \]
Rearranging gives:
\[ \Phi = E - V \]
Substituting the values:
\[ \Phi = 4.969 \, \text{eV} - 0.5 \, \text{eV} \]
\[ \Phi = 4.469 \, \text{eV} \]
### Step 6: Round the work function to appropriate significant figures
The work function can be rounded to:
\[ \Phi \approx 4.5 \, \text{eV} \]
### Final Answer
The work function of the metal is approximately **4.5 eV**.
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