A beam of light has three wavelengths `4144 Å`, `4972 Å` and `6216 Å` with a total instensity of `3.6 xx 10^(-3) Wm^(-2)` equally distributed amongst the three wavelengths. The beam falls normally on an area `1.0 cm^2` of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects on electron. Calculate the number of photo electrons liberated in two seconds.

Explanation

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

### Step 1: Calculate the Energy of Each Photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) = Planck's constant \( = 6.626 \times 10^{-34} \, \text{J s} \)
- \( c \) = Speed of light \( = 3 \times 10^8 \, \text{m/s} \)
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