A beam of electromagnetic radiation comprised of wavelength 620 nm on a perfectly absorbing metal (work function, `phi=2eV)` surface of area `1cm^2` Pressure exerted by the radiation on the surface is `3xx2^(-13)N//cm^3` . If one in `3xx10^3`photons ejects an electron, and total number of electrons ejected in 1 is `1x`, then x is _______
[Given : hc=1240 e Vnm, 1eV=`1.6xx10^(-19)J]`

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Understand the given data
- Wavelength of electromagnetic radiation, \( \lambda = 620 \, \text{nm} = 620 \times 10^{-9} \, \text{m} \)
- Work function of the metal, \( \phi = 2 \, \text{eV} \)
- Pressure exerted by the radiation, \( P = 3 \times 10^{-13} \, \text{N/cm}^2 = 3 \times 10^{-9} \, \text{N/m}^2 \) (since \( 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \))
- The number of photons required to eject one electron, \( N = 3 \times 10^3 \)
- Total number of electrons ejected in 1 second is \( 1x \)

### Step 2: Calculate the energy of a single photon
Using the formula for the energy of a photon:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{620 \times 10^{-9} \, \text{m}} = \frac{1.9878 \times 10^{-25}}{620 \times 10^{-9}} \approx 3.21 \times 10^{-19} \, \text{J} \]

Now, converting this energy into electron volts:
\[ E = \frac{3.21 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 2.01 \, \text{eV} \]

### Step 3: Determine the number of photons incident per second
Using the relationship between pressure, number of photons, and energy:
\[ P = \frac{n_p E}{A} \]
Where:
- \( n_p \) = number of photons incident per second
- \( A \) = area (1 cm² = \( 1 \times 10^{-4} \, \text{m}^2 \))

Rearranging gives:
\[ n_p = \frac{P \cdot A}{E} \]
Substituting the values:
\[ n_p = \frac{(3 \times 10^{-9} \, \text{N/m}^2)(1 \times 10^{-4} \, \text{m}^2)}{3.21 \times 10^{-19} \, \text{J}} \approx \frac{3 \times 10^{-13}}{3.21 \times 10^{-19}} \approx 9.34 \times 10^5 \, \text{photons/s} \]

### Step 4: Calculate the total number of electrons ejected
Given that 1 in \( 3 \times 10^3 \) photons ejects an electron:
\[ \text{Number of electrons ejected} = \frac{n_p}{3 \times 10^3} \]
Substituting the value of \( n_p \):
\[ \text{Number of electrons ejected} = \frac{9.34 \times 10^5}{3 \times 10^3} \approx 311.33 \approx 311 \]

### Step 5: Find the value of \( x \)
Since the total number of electrons ejected in 1 second is \( 1x \):
\[ x = 311 \]

### Final Answer
Thus, the value of \( x \) is **311**.

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