To tackle this problem, we need to break it down into manageable parts. We start by calculating the total power incident on the sodium photo-cell, then determine how much energy is absorbed by the top layers of sodium, and finally, we can estimate the time required for photoelectric emission based on the work function of sodium.
Calculating the Incident Power
The intensity of the light is given as \(10^{-5} \, \text{W/m}^2\), and the surface area of the sodium photo-cell is \(2 \, \text{cm}^2\). First, we need to convert the area into square meters:
- Area in square meters: \(2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2\)
Now, we can calculate the total power (\(P\)) incident on the photo-cell using the formula:
P = Intensity × Area
Substituting the values:
P = \(10^{-5} \, \text{W/m}^2 \times 2 \times 10^{-4} \, \text{m}^2 = 2 \times 10^{-9} \, \text{W}\)
Energy Absorbed by Sodium
Next, we need to find out how much energy is absorbed by the top layers of sodium. The energy absorbed (\(E\)) can be calculated using the power and the time (\(t\)) the light is incident:
E = P × t
However, we need to determine how long it takes for the absorbed energy to equal the work function of sodium. The work function (\(\phi\)) is given as 2 eV. To work in SI units, we convert this energy into joules:
- 1 eV = \(1.6 \times 10^{-19} \, \text{J}\)
- Thus, \(\phi = 2 \, \text{eV} = 2 \times 1.6 \times 10^{-19} \, \text{J} = 3.2 \times 10^{-19} \, \text{J}\)
Estimating Time for Photoelectric Emission
Now, we can set the energy absorbed equal to the work function to find the time:
3.2 × 10^{-19} \, \text{J} = 2 × 10^{-9} \, \text{W} × t
Rearranging this gives:
t = \frac{3.2 × 10^{-19} \, \text{J}}{2 × 10^{-9} \, \text{W}} = 1.6 × 10^{-10} \, \text{s}
Implications of the Result
The calculated time of \(1.6 × 10^{-10} \, \text{s}\) indicates how quickly the photoelectric effect can occur under the given conditions. This rapid response time suggests that sodium is a suitable material for photoelectric applications, as it can efficiently convert light energy into electrical energy almost instantaneously. However, it also highlights the importance of the intensity of light and the material's work function in determining the efficiency of photoelectric emission.
In practical terms, this means that in applications such as photodetectors or solar cells, materials with lower work functions and higher absorption rates can lead to faster and more efficient energy conversion processes.
