Question

The electric field associated with a light wave is given by `E = E_0 sin [(1.57 xx 10^7  "m"^-1)(x - ct)]`. 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 , `E = E_0  sin [(1.57 xx 10^7 "m"^-1)(x - ct)]`

Work function, `phi = 1.9  "eV"`

On comparing the given equation with the standard equation, `E = E_0  sin (kx - wt)`,

we get :- 

`omega = 1.57 xx 10^7 xx c`

Now , frequency,

`v = (1.57 xx 10^7 xx 3 xx 10^8)/(2pi) Hz`

From Einstein's photoelectric equation,

`"eV"_0 = hv - phi`

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

On substituting the respective values, we get :-

`"eV"_0 = 6.63 xx 10^-34 xx (1.57 xx 3 xx 10^15)/(2pi xx 1.6 xx 10^-19) - 1.9  "eV"`

`⇒ "eV"_0 = 3.105 - 1.9 = 1.205  "eV"`

`⇒ V_0 = (1.205 xx 1.6 xx 10^-19)/(1.6 xx 10^-19) = 1.205 V`

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