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Question

The electric field associated with a light wave is given by
E=E0 sin [(1.57×107m-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.

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Solution

Given:
Electric field, E=E0 sin [(1.57×107m-1)(x-ct)]
Work function, ϕ = 1.9 eV
On comparing the given equation with the standard equation, E=E0sinkx-wt, we get:
ω=1.57×107×c
Now, frequency,
v=1.57×107×3×1082π Hz
From Einstein's photoelectric equation,
eV0=hv-ϕ
Here, V0 = stopping potential
e = charge on electron
h = Planck's constant
On substituting the respective values, we get:
eV0=6.63×10-34×1.57×3×10152π×1.6×10-19-1.9 eVeV0=3.105-1.9=1.205 eVV0=1.205×1.6×10-191.6×10-19=1.205 V
Thus, the value of the stopping potential is 1.205 V.

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