Question

The electric field associated with a light wave is given by
$E=E_0\sin \left[\right(1.57\times {10}^7{\mathrm{m}}^{-1}\left)\right(x-ct\left)\right].$
Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1.9 eV.

Answer 1

Given:
$\mathrm{Electric}\mathrm{field},E=E_0\sin \left[\right(1.57\times {10}^7{\mathrm{m}}^{-1}\left)\right(x-ct\left)\right]$
Work function, $\varphi$ = 1.9 eV
On comparing the given equation with the standard equation, $E=E_0\sin \left(kx-wt\right)$, we get:
$\omega =1.57\times {10}^7\times c$
Now, frequency,
$v=\frac{1.57\times {10}^7\times 3\times {10}^8}{2\mathrm{\pi }}\mathrm{Hz}$
From Einstein's photoelectric equation,
$e{\mathrm{V}}_0=hv-\varphi$
Here, V₀ = stopping potential
e = charge on electron
h = Planck's constant
On substituting the respective values, we get:
$$\begin{gathered} e{V}_0=6.63\times {10}^{-34}\times \frac{1.57\times 3\times {10}^{15}}{2\mathrm{\pi }\times 1.6\times {10}^{-19}}-1.9\mathrm{eV} \\ \Rightarrow e{V}_0=3.105-1.9=1.205\mathrm{eV} \\ \Rightarrow V_0=\frac{1.205\times 1.6\times {10}^{-19}}{1.6\times {10}^{-19}}=1.205\mathrm{V} \end{gathered}$$
Thus, the value of the stopping potential is 1.205 V.