Light of frequency 2.5 x 1015 Hz illuminates a

Physics

1 answer:

Nikitich [7]2 years ago

3 0

Answer:

h f = W + KE

Input energy equals work function plus KE of emitted electron

W = 6.63E-34 * 2.5E15 - 6.3 * 1.6E-19

W = 6.63 * 2.5 * 10^-19 - 10.1 * E-19 ev (1ev = 1.6E-19 J)

W = (16.6 - 10.1)E-19 = 6.5E-19 J

h f = 6.5E-19 J for electrons to be emitted with zero KE

f = 6.5E-19 / 6.63E-34 = .98E-15 / sec = 9.8E-14 / sec (threshold)

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