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प्रश्न
The electric field at a point associated with a light wave is `E = (100 "Vm"^-1) sin [(3.0 xx 10^15 "s"^-1)t] sin [(6.0 xx 10^15 "s"^-1)t]`.If this light falls on a metal surface with a work function of 2.0 eV, what will be the maximum kinetic energy of the photoelectrons?
(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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उत्तर
Given :-
`E = 100 sin [(3 xx 10^-15 "s"^-1)t] sin [(6 xx 10^-15 "s"^-1)t]`
= `100 xx 1/2 cos [(9 xx 10^15 "s"^-1)t] - cos[(3 xx 10^15 "s"^-1)t]`
The values of angular frequency `ω` are `9 xx 10^15` and `3 xx 10^15`.
Work function of the metal surface, `phi = 2 "eV"`
Maximum frequency,
`v = ω_(max)/(2pi) = (9 xx 10^15)/(2pi) Hz`
From Einstein's photoelectric equation, kinetic energy,
K = hv - `phi`
`⇒ K = 6.63 xx 10^-34 xx (9 xx 10^15)/(2pi) xx 1/(1.6 xx 10^-19) - 2 "eV"`
⇒ K = 3.938 eV
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संबंधित प्रश्न
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In \[E=h\nu\], what does \[\nu\] represent?
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