NCERT Exercise 2.51 - Structure of Atom

Question: For caesium W0 = 1.9 eV, find threshold wavelength/frequency and photoelectron K.E./speed at 500 nm.

Answer: lambda0 = 652.46 nm; nu0 = 4.598 x 10^14 s^-1; K.E. = 9.29 x 10^-20 J; speed = 4.516 x 10^5 m s^-1.

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