A beam of light has two wavelengths `4,972Å and 6,216Å ` with a total intensity of `3.6xx10^(-3) W m^(-2)` equally distributed among the two wavelengths . The beam falls normally on an area of `1 cm^2` of clean metallic surface of work function 2.3 eV . Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in 2 s is approximately

Options

A. `6xx10^(11)`
B. `9xx10^(11)`
C. `11xx10^(11)`
D. `15xx10^(11)`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Determine the Intensity for Each Wavelength
The total intensity of the beam is given as \(3.6 \times 10^{-3} \, \text{W/m}^2\). Since the intensity is equally distributed among the two wavelengths, we can calculate the intensity for each wavelength as follows:

\[ I_1 = I_2 = \frac{3.6 \times 10^{-3}}{2} = 1.8 \times 10^{-3} \, \text{W/m}^2 \]

### Step 2: Calculate the Energy of Photons for Each Wavelength
The energy of a photon can be calculated using the formula:

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

Where:
- \(h = 6.63 \times 10^{-34} \, \text{J s}\) (Planck's constant)
- \(c = 3 \times 10^8 \, \text{m/s}\) (speed of light)
- \(\lambda\) is the wavelength in meters.

First, we convert the wavelengths from angstroms to meters:
- \(\lambda_1 = 4972 \, \text{Å} = 4972 \times 10^{-10} \, \text{m}\)
- \(\lambda_2 = 6216 \, \text{Å} = 6216 \times 10^{-10} \, \text{m}\)

Now we can calculate the energy for each wavelength:

For \(\lambda_1\):
\[ E_1 = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{4972 \times 10^{-10}} = 2.49 \, \text{eV} \]

For \(\lambda_2\):
\[ E_2 = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{6216 \times 10^{-10}} = 1.994 \, \text{eV} \]

### Step 3: Determine Which Wavelength Can Eject Electrons
The work function (\(\phi\)) of the metallic surface is given as \(2.3 \, \text{eV}\). We compare the energies of the photons with the work function:

- For \(\lambda_1\): \(E_1 = 2.49 \, \text{eV} > 2.3 \, \text{eV}\) (can eject electrons)
- For \(\lambda_2\): \(E_2 = 1.994 \, \text{eV} < 2.3 \, \text{eV}\) (cannot eject electrons)

Thus, only photons of wavelength \(\lambda_1\) can eject electrons.

### Step 4: Calculate the Number of Photons Incident on the Surface
The number of photons incident on the surface can be calculated using the formula:

\[ N = \frac{I \cdot A \cdot t}{E} \]

Where:
- \(I = 1.8 \times 10^{-3} \, \text{W/m}^2\) (intensity for \(\lambda_1\))
- \(A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2\) (area)
- \(t = 2 \, \text{s}\) (time)
- \(E = 2.49 \, \text{eV} = 2.49 \times 1.6 \times 10^{-19} \, \text{J}\) (energy of a photon in joules)

Calculating \(E\):
\[ E = 2.49 \times 1.6 \times 10^{-19} = 3.984 \times 10^{-19} \, \text{J} \]

Now substituting into the formula for \(N\):
\[ N = \frac{(1.8 \times 10^{-3}) \cdot (1 \times 10^{-4}) \cdot 2}{3.984 \times 10^{-19}} \]

Calculating \(N\):
\[ N = \frac{3.6 \times 10^{-7}}{3.984 \times 10^{-19}} \approx 9.04 \times 10^{11} \]

### Step 5: Conclusion
The total number of photoelectrons liberated in 2 seconds is approximately \(9 \times 10^{11}\).

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