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The electric field associated with a light wave is given by E = E base 0 sin [(1.57 × 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.

The electric field associated with a light wave is given by
E = E base 0 sin [(1.57 × 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.

Grade:10

2 Answers

Kevin Nash
askIITians Faculty 332 Points
7 years ago
Sol. E = E base 0 sin[(1.57 * 10^7 m^-1) (x - ct)] W = 1.57 * 10^7 * C ⇒ f = 1.57 * 10^7 *3 * 10^8/2π Hz W base 0 = 1.9 ev Now eV base 0 = hv – W base 0 = 4.14 * 10^15 * 1.57 * 3 * 10^15/2π – 1.9 ev = 3.105 – 1.9 = 1.205 ev So, V base 0 = 1.205 * 1.6 *10^-19/1.6 * 10^-19 = 1.205 V.
Rohit
15 Points
3 years ago
As we know
In general form Ef=Ef°sin(kx-wt)
Where w is angular frequency and other notation have specific meaning 
And compare it with given question equation we get w=1.57*10^7*c
And we know very well E=hw/2*3.14
And kinetic energy =E- work function
K.E.=3.105-1.9=1.205ev
AND eV=K.E.
V=K.E.÷e=1.205ev÷e=1.205

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