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.

Explanation

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)`