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 given problem, we will break it down into two parts:

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

1. **Understanding the Work Function**: The work function (φ) of cesium is given as 2.14 eV. To find the threshold frequency (ν₀), we need to convert this energy into joules. The conversion factor is 1 eV = 1.6 × 10^(-19) J.

\[
φ = 2.14 \, \text{eV} = 2.14 \times 1.6 \times 10^{-19} \, \text{J} = 3.424 \times 10^{-19} \, \text{J}
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