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29 January, 20:29

When ultraviolet light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 ev. part a what is the maximum kinetic energy of the photoelectrons when light of wavelength 295 nm falls on the same surface? emax = ?

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  1. 29 January, 21:07
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    A general relationship between Kinetic Energy (KE) and wavelength (y) is that KE is inversely proportional to square of y. That is,

    KE α 1/y^2 = > KE = k/y^2 where k is a constant of proportionality.

    For KE = 1.10 ev, and y = 400 nm;

    k = KE*y^2 = 1.10*400 = 176,000 ev. nm^2

    For y = 295 nm,

    KE = k/y^2 = 176,000/295^2 = 2.02 ev.

    Therefore, e max = 2.02 ev
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