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UV light of wavelength 1800Å is incident on a lithium surface whose threshold wavelength 4965Å. Determine the maximum energy of the electron emitted. |
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Answer» SOLUTION :`{:("Maximum kinetic energy of electron,",,LAMBDA = 1800 XX 10^(-10)m),(,,lambda_(0) = 4965 xx 10^(10)m),(K_(max) = hc((1)/(lambda)-(1)/(lambda_(0))),,h = 6.6 xx 10^(-34)Js),(,,c = 3 xx 10^(8) ms^(-1)):}` `= (6.6 xx 10^(-34) xx 3 xx 10^(8))/(10^(-10)) xx ((1)/(1800) - (1)/(4965))` `= 19.8 xx 10^(-16) xx (3165)/(8937 xx 10^(3)) = 7.01208 xx 10^(-19)J` `= (7.01208 xx 10^(-19))/(1.6 xx 10^(-19)) = 4.38 eV` `K_(max) = 4.40 eV` |
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