Physics

Problem

The electric field of light wave is given as

\vec{E} = 10^3 \cos\left(\frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14} t\right) \hat{x} \frac{N}{C}

This light falls on a metal plate of work function

2\text{eV}

. The stopping potential of the photo-electrons is:

\text{Given, } E (\text{in eV}) = \frac{12375}{\lambda (\text{in } \text{\AA})}

[Main 9 April 2019 (I)] (a)

2.0\text{V}

(b)

0.72\text{V}

(c)

0.48\text{V}

(d)

2.48\text{V}