The threshold wave length of photosensitive metal is 5000 A find the K.E of the photo electrons emmited by it pt radiation of wave length 4000A incident on it explain in ev
Binita , 7 Years ago
Grade 12th pass
1 Answers
vipul singh
Last Activity: 7 Years ago
Threshold wavelength( λ¹) = 5000A ∴ threshold energy(ф)= hc/λ¹= 12400/5000 ev = 2.48 ev Wavelength of radiation(λ) = 4000A ∴ energy of radiation(E)= hc/λ = 12400/4000 ev = 3.1 EV Since E = ф + K.E.(max.) 3.1 = 2.48 + K.E.(max) K.E.(max) = 0.62 ev
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