A Chemistry question

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Oct 8th, 2014

electron's mass = 9.11 * 10^-31 kg
Energy of incident photon = threshold energy + KE of electron
KE = Energy of incident photon - threshold energy
Energy of incident photon = hf
= hc/λ
KE = (1/2)mv^2

KE = hc/λ - threshold energy
(1/2)mv^2 = (6.626 * 10^-34 Js)(3.0 * 10^8 m/s)/(67.2 nm) - 7.77 * 10^-19 J
(1/2)mv^2 = (6.626 * 10^-34 Js)(3.0 * 10^8 m/s)/(67.2 * 10^-9 m) - 7.77 * 10^-19 J
(1/2)mv^2 = 2.18103571 * 10^-18 J
v = √ (2*(6.626 * 10^-34 Js)(3.0 * 10^8 m/s)/(67.2 * 10^-9 m) - 7.77 * 10^-19 J)/m)
= √ (2*(6.626 * 10^-34 Js)(3.0 * 10^8 m/s)/(67.2 * 10^-9 m) - 7.77 * 10^-19 J)/(9.11 * 10^-31 kg))
= 2375106.78 m/s
= 2.37510678 * 10^6 m/s

Since 2.37510678 * 10^6 m/s < 0.1c, the electron's speed is not relativistic and thus, distortions due to special relativity do not have to be taken into account.

Rounding to three sig figs, the electron's speed is 2.38 * 10^6 m/s.

Its kinetic energy is 2.18 * 10^-18 J.

Oct 8th, 2014

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