Asked by daniela
You have three metals - lithium, aluminum, and mercury. Their work functions are: 2.30, 4.10eV, and 4.5eV respectively. Light with a frequency of 1 x 10^15 Hz is incident on all three metals. Determine (a) which metals will emit electrons and (b)the maximum kinetic energy for those that exhibit the effect.
Answers
Answered by
drwls
(a) The photon energy is
E = h*f = 6.62*10^-34 * 1*10^15 = 6.62*10^-19 J
Since 1 eV = 1.602*10^-19 J,
E = 4.13 eV.
Lithium and aluminum have work functions lower than that, and will emit photoelectrons
(b) The maximum photoelectron kinetic energy equals the photon energy (in eV) minus the work function
E = h*f = 6.62*10^-34 * 1*10^15 = 6.62*10^-19 J
Since 1 eV = 1.602*10^-19 J,
E = 4.13 eV.
Lithium and aluminum have work functions lower than that, and will emit photoelectrons
(b) The maximum photoelectron kinetic energy equals the photon energy (in eV) minus the work function
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