Question

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

(Use h = 6.63 × 10^-34J-s = 4.14 × 10^-15 eV-s, c = 3 × 10⁸ m/s and me = 9.1 × 10^-31kg)

Answer 1

Given :-

Electric field , ${\displaystyle E=E_0 \sin \left[(1.57\times {10}^7{\text{m}}^{-1})(x-ct)\right]}$

Work function, ${\displaystyle \varphi =1.9 \text{eV}}$

On comparing the given equation with the standard equation, ${\displaystyle E=E_0 \sin (kx-wt)}$,

we get :- 

${\displaystyle \omega =1.57\times {10}^7\times c}$

Now , frequency,

${\displaystyle v=\frac{1.57\times {10}^7\times 3\times {10}^8}{2\pi }Hz}$

From Einstein's photoelectric equation,

${\displaystyle {\text{eV}}_0=hv-\varphi }$

Here, V0 = stopping potential
           e = charge on electron
           h = Planck's constant

On substituting the respective values, we get :-

${\displaystyle {\text{eV}}_0=6.63\times {10}^{-34}\times \frac{1.57\times 3\times {10}^{15}}{2\pi \times 1.6\times {10}^{-19}}-1.9 \text{eV}}$

${\displaystyle \Rightarrow {\text{eV}}_0=3.105-1.9=1.205 \text{eV}}$

${\displaystyle \Rightarrow V_0=\frac{1.205\times 1.6\times {10}^{-19}}{1.6\times {10}^{-19}}=1.205V}$

Thus, the value of the stopping potential is 1.205 V.