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Karnataka Board PUCPUC Science Class 11

The Electric Field Associated with a Light Wave is Given - Physics

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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)

Sum
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Solution

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.

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Experimental Study of Photoelectric Effect
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Chapter 20: Photoelectric Effect and Wave-Particle Duality - Exercises [Page 365]

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HC Verma Concepts of Physics Vol. 2 [English] Class 11 and 12
Chapter 20 Photoelectric Effect and Wave-Particle Duality
Exercises | Q 21 | Page 365

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