A light of wavelength 600 nm is incident on a metal surface. When light of wavelength 400 nm is incident, the maximum kinetic energy of the emitted photoelectrons is doubled. The work function of the metals is

Options

A. 1.03 eV
B. 2.11 eV
C. 4.14 eV
D. 2.43 eV

Answer

A

Explanation

(a) : Here, `lamda=600nm, lamda^(')=400nm,K_(max)^(')=2K_(max)`
According to Einstein's photoelectic equation
`K_(max)=(hc)/(lamda)-phi_(0)" "(i)`
and `2K_(max)=(hc)/(lamda^('))-phi_(0)" "(ii)`
Dividing (ii) by (i) we get
`2=(((hc)/(lamda^(')))-phi_(0))/(((hc)/(lamda))-phi_(0))or(2hc)/(lamda)-2phi_(0)=(hc)/(lamda^('))-phi_(0)`
`or hc((2)/(lamda)-(1)/(lamda^(')))=phi_(0)` (Take hc = 1240 eV nm)
`:.phi_(0)=1240eVnm((2)/(600nm)-(1)/(400nm))=1.03eV`