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Juliakg1 Juliakg1
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11 years ago
The photoelectric binding energy of cesium is 183.7 kJ/mol. Light having a wavelength of
2.4 × 10?7 m falls upon a cesium surface in an evacuated tube. Calculate the minimum deBroglie
wavelength of the emitted photoelectrons.
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wrote...
11 years ago
for photoelectric effects/experiments

E (incoming photon) = B.E. + K.E. (emitted electron)

E (photon incoming) = hc / lambda_light (of the photon incoming)
where h = 6.63 * 10^-34  J*s
and c = 3.00 * 10^ m/s
(lambda of the photon incoming = c / "nu")
KE (emitted electron) =.5mv^2
and lamdba_DeBroglie (emitted electron) = h / mv
mass of electron = 9.11 * 10 ^ -34 g (you may have to convert this)

given Binding Energy and lambda_light (for the photon incoming), solve for K.E. (emitted electron), then v (emitted electron), then lambda (emitted electron). ignore "minimum" here.
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