Find the maximum velocity of photoelectrons emitted by radiation of frequency `3 xx 10^(15)` Hz from a photoelectric surface having a work function of 4.0 eV. Given `h=6.6 xx 10^(-34)` Js and 1 eV = `1.6 xx 10^(-19)` J.

Explanation

To find the maximum velocity of photoelectrons emitted by radiation of frequency \(3 \times 10^{15}\) Hz from a photoelectric surface with a work function of 4.0 eV, we can follow these steps:

### Step 1: Convert the work function from eV to Joules
The work function \(\phi\) is given in electron volts (eV). We need to convert it to Joules using the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\).

\[ \phi = 4.0 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 6.4 \times 10^{-19} \text{ J} \]
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