A metal plate is placed 5m from a monchromatic ligth
soure whose power output is `10^(-3)`W. Consider that a given ejected
photoelectron may collect its energy from a circular area of the plate as large as
ten atomic diameters `(10^(-9m)` in radius. The energy required to remove an electron through the metal surface is about 5.0 eV. Assuming light to be a
wave, how long would it take for such a 'target' to soak up this much energy
from such a light source.

Explanation

To solve the problem step by step, we need to calculate the time it takes for a photoelectron to absorb enough energy from the monochromatic light source to be ejected from the metal plate.

### Step 1: Calculate the area from which the photoelectron can collect energy

The radius of the circular area from which the photoelectron can collect energy is given as \(10^{-9}\) m. The area \(A\) of this circle can be calculated using the formula:

\[ A = \pi r^2 \]

Substituting \(r = 10^{-9}\) m:

\[ A = \pi (10^{-9})^2 = \pi \times 10^{-18} \text{ m}^2 \]

### Step 2: Calculate the intensity of the light at the metal plate

The power \(P\) of the light source is given as \(10^{-3}\) W. The intensity \(I\) of the light can be calculated using the formula:

\[ I = \frac{P}{A_{\text{sphere}}} \]

Where \(A_{\text{sphere}}\) is the surface area of a sphere with a radius of 5 m:

\[ A_{\text{sphere}} = 4\pi r^2 = 4\pi (5)^2 = 100\pi \text{ m}^2 \]

Now, substituting the values into the intensity formula:

\[ I = \frac{10^{-3}}{100\pi} = \frac{10^{-3}}{100\pi} \text{ W/m}^2 \]

### Step 3: Calculate the power incident on the area from which the photoelectron collects energy

The power \(P_{\text{incident}}\) incident on the area \(A\) can be calculated as:

\[ P_{\text{incident}} = I \times A = \left(\frac{10^{-3}}{100\pi}\right) \times \left(\pi \times 10^{-18}\right) \]

This simplifies to:

\[ P_{\text{incident}} = \frac{10^{-3} \times 10^{-18}}{100} = 10^{-21} \text{ W} \]

### Step 4: Convert the energy required to eject an electron from eV to Joules

The energy required to remove an electron is given as 5.0 eV. To convert this to Joules, we use the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\):

\[ E = 5.0 \times 1.6 \times 10^{-19} = 8.0 \times 10^{-19} \text{ J} \]

### Step 5: Calculate the time required to absorb enough energy

The time \(t\) required to absorb this energy can be calculated using the formula:

\[ t = \frac{E}{P_{\text{incident}}} \]

Substituting the values:

\[ t = \frac{8.0 \times 10^{-19}}{10^{-21}} = 8.0 \times 10^{2} \text{ seconds} \]

### Step 6: Convert time from seconds to hours

To convert seconds into hours, we divide by 3600 (the number of seconds in an hour):

\[ t = \frac{8.0 \times 10^{2}}{3600} \approx 0.222 \text{ hours} \approx 13.33 \text{ minutes} \]

### Final Answer

It would take approximately **13.33 minutes** for the target to absorb enough energy to eject a photoelectron.

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