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Question
Threshold frequency, ν0 is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency 1.0 × 1015 s–1 was allowed to hit a metal surface, an electron having 1.988 × 10–19 J of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.
Solution
For the emission of electrons from metal the frequency of the striking light should be higher than that of its threshold frequency.
We have `hv = hv_0 + K.E`
`v_0 = (v - K.E)/h` ......(1)
Given, `v = 10^15 s^-1`, K.E = `1.988 xx 10^-19` J
Thus (1) gives
`v_0 = 7 xx 10^14 s^-1`
Given, `lambda` = 600 nm
`v = c/lambda = 5 xx 10^14 s^-1`
As `v_0 > v`, thus the electrons do not emit.
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