What is the maximum speed of an electron emitted from a metal surface with a threshold frequency of 5.706*10^(14) by light with a wavelength of 350nm?
In order to determine the maximum speed of an emitted electron, we must first find it's kinetic energy from the energy of an incident photon and the work function of the metal surface (the minimum energy required for an electron to escape the metal surface). These are related using the equation below. E(total energy provided by photon)=Φ+EkThe energy of a single photon can be calculated from Planck's Equation, E=hf, where E is energy, h is Planck's constant (6.6310^(-34)Js) and f is frequency. In this case we don't have the frequency of the incident, but the wavelength instead, therefore in order to obtain the frequency the equation f=c/lambda can be used, where f is frequency, c is the speed of light (3.0010^(8)m/s) and lambda is the wavelength.Then use the given values to find the energy of the photon. => f=3.0010^(8)/(35010^(-9))=8.5710^(14) Hz=> E=6.6310^(-34)8.5710^(14) = 5.6810^(-19) JThe work function, Φ, can also be calculated from the threshold frequency using Planck's equation. => Φ=6.6310^(-34)5.70610^(14)= 3.7810^(-19) JTherefore Ek(max)=E-Φ= (5.68-3.78)10^(-19) J = 1.9010^(-19) JRearranging Ek(max)=(1/2)mv2 for v, gives vmax=(2Ek(max)/m)^(1/2) = 6.4610^(5) m/s
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