The work function of cesium is 2.14 eV. Find (a) the threshold frequency for cesium, and (b) the wavelength of the incident light if the photo current is brought to zero by a stopping potential 0.60 V. Given `h=6.63xx10^(-34)Js`.

Explanation

To solve the problem step by step, we will break it down into two parts as requested.

### Part (a): Finding the Threshold Frequency for Cesium

1. **Identify the Work Function**:
The work function (W) of cesium is given as 2.14 eV. We need to convert this into joules.
\[
W = 2.14 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.424 \times 10^{-19} \, \text{J}
...