A beam of electromagnetic radiation of intensity `12.8 xx 10^(-5) W//cm^(2)` is comprised of wavelength, `lambda = 155 nm` . It falls normally on a metal (work function `phi = 2eV`) of surfacce area of the ` 1 cm^(2)`. If one in `10^(3)` photons ejects total number of electrons ejected in 1 s in `10^(x)`. (hc = 1240 eVnm) `1eV = 1.6 xx 10^(-19)`J =, then x is _____.

Explanation

To solve the problem step by step, we will follow the outlined process:

### Step 1: Understand the given data
We have the following information:
- Intensity of the radiation: \( I = 12.8 \times 10^{-5} \, \text{W/cm}^2 \)
- Wavelength of the radiation: \( \lambda = 155 \, \text{nm} \)
- Work function of the metal: \( \phi = 2 \, \text{eV} \)
- Surface area of the metal: \( A = 1 \, \text{cm}^2 \)
- \( hc = 1240 \, \text{eV nm} \)
- \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \)

### Step 2: Calculate the energy of a single photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Substituting the values:
\[ E = \frac{1240 \, \text{eV nm}}{155 \, \text{nm}} = 8 \, \text{eV} \]

### Step 3: Check if the photoelectric effect occurs
Since the energy of the photon \( E = 8 \, \text{eV} \) is greater than the work function \( \phi = 2 \, \text{eV} \), the photoelectric effect will occur.

### Step 4: Calculate the power incident on the surface
The power incident on the surface area can be calculated using:
\[ P = I \times A \]
Substituting the values:
\[ P = 12.8 \times 10^{-5} \, \text{W/cm}^2 \times 1 \, \text{cm}^2 = 12.8 \times 10^{-5} \, \text{W} \]

### Step 5: Convert power to energy in electron volts
To convert the power from watts to electron volts, we need to convert joules to electron volts. Since \( 1 \, \text{W} = 1 \, \text{J/s} \), we can express the power in terms of energy per second:
\[ P = 12.8 \times 10^{-5} \, \text{W} = 12.8 \times 10^{-5} \, \text{J/s} \]

Now converting joules to electron volts:
\[ P = \frac{12.8 \times 10^{-5}}{1.6 \times 10^{-19}} \, \text{eV/s} \]

### Step 6: Calculate the number of photons incident per second
The number of photons \( n \) incident per second can be calculated using:
\[ n = \frac{P}{E} \]
Substituting the values:
\[ n = \frac{12.8 \times 10^{-5} \, \text{W}}{8 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV}} = \frac{12.8 \times 10^{-5}}{1.28 \times 10^{-18}} = 10^{14} \, \text{photons/s} \]

### Step 7: Calculate the number of electrons ejected
Given that 1 in \( 10^3 \) photons ejects 1 electron, the total number of electrons \( N \) ejected in one second can be calculated as:
\[ N = \frac{n}{10^3} = \frac{10^{14}}{10^3} = 10^{11} \, \text{electrons} \]

### Step 8: Determine the value of \( x \)
Since \( N = 10^{11} \), we can express this as \( 10^x \) where \( x = 11 \).

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{11} \]