When a metal surface is illuminated with light of wavelength 437 nm, the stopping potential for photoelectrons is 1.67 V. (c = 3.00 × 108 m/s, h = 6.626 × 10-34 J · s, = - 1.60 × 10-19 C, 1 eV = 1.60 × 10-19 J, el = 9.11 × 10-31 kg)
What's the maximum speed of the emitted electron?
Thank you!
What's the maximum speed of the emitted electron?
Thank you!
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the energy of the photon is h c/L = 6.63x10^-34Js x 3x10^8m/s / 4.37x10^-7 m = 4.55x10^-19J
1.67 eV of this is used to eject the electron from the surface; this is equivalent to
1.6 x10^-19J/eV x 1.67eV = 2.67 x 10^-19J
this means that 4.55 x 10^-19J - 2.67x10^-19J= 1.88x10^-19 J go into the KE of the ejected electron
KE = 1/2 m v^2 so that v=Sqrt[2 KE/m] Sqrt[2 x 1.88x10^-19J/9.11 x 10^-31kg] = 6.42x10^5m/s
1.67 eV of this is used to eject the electron from the surface; this is equivalent to
1.6 x10^-19J/eV x 1.67eV = 2.67 x 10^-19J
this means that 4.55 x 10^-19J - 2.67x10^-19J= 1.88x10^-19 J go into the KE of the ejected electron
KE = 1/2 m v^2 so that v=Sqrt[2 KE/m] Sqrt[2 x 1.88x10^-19J/9.11 x 10^-31kg] = 6.42x10^5m/s