The maximum velcities of the photoelectrons ejected are v and 2v for incident light of wavelength 400nm and 250nm on a metal surface respectively. Calculate the work function of the metal. Given `h=6.62xx10^(-34)Js and c=3xx10^(8) ms^(-1)`

Answer

`2.488 eV`

Explanation

According to Eistein's photoelectric equation
`1/2mv^(2)=(hc)/lambda-phi_(0)`
case (i) `1/2mv^(2)=(hc)/(lambda_(1))-phi_(0)`
Case (ii) `1/2m (2v^(2))=(hc)/(lambda_(2))-phi_(0)`
or `4(1/2mv^(2))=(hc)/(lambda_(2))-phi_(0)`
or `4((hc)/(lambda_(1))-phi_(0))=(hc)/(lambda_(2))-phi_(0) or 3 phi_(0)=(4hc)/(lambda_(1))-(hc)/(lambda_(2))`
or `phi_(0)=(hc)/3[4/(lambda_(1))-1/(lambda_(2))]`
`=(6.63xx10^(-34)xx3xx10^(8))/3xx[4/(400xx10^(-9))-1/(250xx10^(-9))]`
`6.63xx10^(-26)[1/(10^(-7))-1/(2.5xx10^(-7))]`
`=(6.63xx10^(-26))/(10^(-7))[1-10/25]=6.63xx10^(-19)xx3/5`
`=3.978xx10^(-19)J=(3.978xx10^(-19))/(1.6xx10^(-19))eV`
`=2.486 eV`