The work function for sodium metal is `2.46eV`. Determine the cutoff wavelength for sodium? Strategy: The cutoff frequency `v_(0)` is related to the work function through the relation `v_(0) = W_(0)//h`. This corresponds to a cutoff wavelength of `lambda_(0) = (C )/(v_(0)) = (C)/(W_(0)//h) = (hc)/(W_(0))`. wavelength grater than `lambda_(0)` for a metal with work function `W_(0)` produce no photoelectric effect.

The work function for sodium metal is `2.46eV`. Determine the cutoff w

1.	The work function for sodium metal is `2.46eV`. Determine the cutoff wavelength for sodium? Strategy: The cutoff frequency `v_(0)` is related to the work function through the relation `v_(0) = W_(0)//h`. This corresponds to a cutoff wavelength of `lambda_(0) = (C )/(v_(0)) = (C)/(W_(0)//h) = (hc)/(W_(0))`. wavelength grater than `lambda_(0)` for a metal with work function `W_(0)` produce no photoelectric effect.
Answer» Step `1.` First convert work function `W_(0)` from electron volts to joules. `W_(0) = 2.46eV` `=(2.46eV) ((1.6xx10^(-19)J)/(1eV))` `= 3.9 xx 10^(-19)J` Step `2.` Now solve the equation for the cutoff wavelength `lambda_(0) = (hc)/(W_(0))` `=((6.6xx10^(-34)Js)(3.0xx10^(8)ms^(-1)))/((3.9xx10^(-19)J))` `= 5.1 xx 10^(-7)m` `= 510 nm (1nm = 10^(-9)m)`

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