The cathode of a photocell is irradiated with light of wavelength `3000 Å`. The current through the cell is found to stop when the potential of the plate is 2.1 V just below the cathode. Calculate the work function of the cathode.

Answer

To solve the problem, we will use the principles of the photoelectric effect and Einstein's equation. Let's break down the steps:

### Step 1: Convert the wavelength to frequency
The wavelength of the light is given as \( \lambda = 3000 \, \text{Å} \). We need to convert this to meters:
\[ \lambda = 3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} = 3 \times 10^{-7} \, \text{m} \]
Now, we can calculate the frequency \( \nu \) using the equation:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light (\( c \approx 3 \times 10^8 \, \text{m/s} \)):
\[ \nu = \frac{3 \times 10^8 \, \text{m/s}}{3 \times 10^{-7} \, \text{m}} = 10^{15} \, \text{Hz} \]

### Step 2: Calculate the energy of the photons
Using the frequency, we can calculate the energy of the photons \( E \) using the equation:
\[ E = h \nu \]
where \( h \) is Planck's constant (\( h \approx 6.626 \times 10^{-34} \, \text{Js} \)):
\[ E = 6.626 \times 10^{-34} \, \text{Js} \times 10^{15} \, \text{Hz} = 6.626 \times 10^{-19} \, \text{J} \]

### Step 3: Convert the energy from Joules to electron volts
To convert the energy from Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{6.626 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 4.14 \, \text{eV} \]

### Step 4: Use the stopping potential to find the work function
The stopping potential \( V_0 \) is given as \( 2.1 \, \text{V} \). The kinetic energy of the emitted electrons when they just stop is equal to the charge of the electron multiplied by the stopping potential:
\[ K.E. = eV_0 = 1.6 \times 10^{-19} \, \text{C} \times 2.1 \, \text{V} = 3.36 \times 10^{-19} \, \text{J} \]
Converting this to electron volts:
\[ K.E. = 2.1 \, \text{eV} \]

### Step 5: Calculate the work function
Using Einstein's photoelectric equation:
\[ E = \phi + K.E. \]
where \( \phi \) is the work function. Rearranging gives:
\[ \phi = E - K.E. \]
Substituting the values:
\[ \phi = 4.14 \, \text{eV} - 2.1 \, \text{eV} = 2.04 \, \text{eV} \]

### Final Answer
The work function of the cathode is:
\[ \phi \approx 2.04 \, \text{eV} \]

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