What wavelength of radiation must be used to ejct eltrons with a velocity of 1600 km/s?
The work function for chromium metal is 4.37 eV. What wavelength of radiation must be used to eject electrons with a velocity of 1600 km/s?
Answer in units of nm.
I am a little lost on this problem, mostly on what values are required to fill out an equation that allow me to solve for what i need. I would really appreciate some step by step instructions, thanks.
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When incoming light is incident on the surface of a metal, each photon that hits an electron in a atom of the metal will excite this electron. If the incoming photon will have enough energy (`E_(ph) =h*F` ), the electron will be extracted from the atom. The threshold energy of the incoming photon for this effect to be produced (extraction of electron from the atom) is named the work function of the metal (`W` ). If the incoming photon has the energy higher than the work function of the metal the rest of the photon energy will be found in the kinetic energy of the ejected electron (`E_k =(mv^2)/2` ). This effect is called the external photoelectric effect and as said above its law is:
`E_(ph) = W +E_k` or equivalent
`h*F = W +(mv^2)/2`
`h*c/lambda = W +(m*v^2)/2`
`lambda = h*c/(W+(m*v^2)/2)`
`lambda = (6.626*10^-34*3*10^8)/(4.37*1.6*10^-19 +0.5*9.1*10^-31*(1600*10^3)^2) =1.066*10^-7 m=106.6 nm`
The wavelength of the incoming light need to be 106.6 nm.
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