A meacury arec lamp provides `0.1` watt of ultra-violet radiation at a wavelength of `lambda=2537 Å` only. The photo tube (cathode of photo electric device) consists of potessium and has an effective area of `4 cm^(2)`. The cathode is located at a distance of `1 m` from the radiation source. The work funcation for potassium is `phi_(0) = 2.22 eV`.
According to classical theory, the radiation from arc lamp spreads out uniformaly in space as spherical wave. What time of expource to the radiation should be required for a potassium atom (radius `2Å`) ub tge cathode to accumulate sufficient enerfy to eject a photo-electron?

Options

A. `352` second
B. `176` second
C. `704` seconds
D. None time lag

Explanation

To solve the problem, we need to determine the time of exposure required for a potassium atom in the cathode to accumulate sufficient energy to eject a photoelectron. We'll follow these steps:

### Step 1: Calculate the energy of the incident radiation

The energy of a photon can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda = 2537 \, \text{Å} = 2537 \times 10^{-10} \, \text{m} \)

Substituting the values:

\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{2537 \times 10^{-10} \, \text{m}} \]

Calculating this gives:

\[ E \approx 7.83 \times 10^{-19} \, \text{J} \]

### Step 2: Convert energy from Joules to electron volts

To convert the energy from joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):

\[ E \text{ (in eV)} = \frac{7.83 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 4.89 \, \text{eV} \]

### Step 3: Compare the energy of the incident radiation with the work function

The work function for potassium is given as \( \phi_0 = 2.22 \, \text{eV} \). Since the energy of the incident radiation \( (4.89 \, \text{eV}) \) is greater than the work function \( (2.22 \, \text{eV}) \), a photoelectron will be ejected.

### Step 4: Calculate the intensity of the radiation at the cathode

The intensity \( I \) of the radiation can be calculated using the power and the area:

\[ I = \frac{P}{A} \]

Where:
- \( P = 0.1 \, \text{W} \) (power of the lamp)
- \( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)

Substituting the values:

\[ I = \frac{0.1 \, \text{W}}{4 \times 10^{-4} \, \text{m}^2} = 250 \, \text{W/m}^2 \]

### Step 5: Calculate the energy incident on the cathode per second

The energy incident on the cathode per second can be calculated as:

\[ E_{\text{incident}} = I \times A \]

Substituting the values:

\[ E_{\text{incident}} = 250 \, \text{W/m}^2 \times 4 \times 10^{-4} \, \text{m}^2 = 0.1 \, \text{W} = 0.1 \, \text{J/s} \]

### Step 6: Calculate the time of exposure required

To find the time \( t \) required for a potassium atom to accumulate sufficient energy, we need to find the number of photons required to provide the work function energy:

\[ \text{Number of photons} = \frac{\phi_0}{E} \]

Substituting the values:

\[ \text{Number of photons} = \frac{2.22 \, \text{eV}}{4.89 \, \text{eV}} \approx 0.453 \]

Since we need at least one photon to eject a photoelectron, we can consider it as 1 photon.

The energy required is \( 2.22 \, \text{eV} = 2.22 \times 1.6 \times 10^{-19} \, \text{J} = 3.552 \times 10^{-19} \, \text{J} \).

Now, we can find the time of exposure:

\[ t = \frac{\text{Energy required}}{\text{Power}} = \frac{3.552 \times 10^{-19} \, \text{J}}{0.1 \, \text{W}} = 3.552 \times 10^{-18} \, \text{s} \]

### Final Answer

The time of exposure required for a potassium atom to accumulate sufficient energy to eject a photoelectron is approximately \( 3.552 \times 10^{-18} \, \text{s} \).

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