What amount of energy sould be added to an electron to reduce its de Broglie wavelength `lamda_1=550nm` incident on it, causing the ejection of photoelectrons for which the stopping potential is `V_(s1)=0.19V`. If the radiation of wavelength `lamda_2=190nm` is now incident on the surface, (a) calculate the stopping potential `V_(S2)`, (b) the work function of the surface, and (c) the threshold frequency for the surface.

Explanation

To solve the problem step by step, we will follow the outlined approach in the video transcript and perform the necessary calculations.

### Step 1: Calculate the Maximum Kinetic Energy (Kmax) for the first wavelength (λ1 = 550 nm)

The stopping potential (Vs1) is given as 0.19 V. The maximum kinetic energy (Kmax) of the ejected photoelectrons can be calculated using the formula:

\[
K_{\text{max}} = eV_{s1}
...