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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)
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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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