Answer
Work function of the material is $0.44eV$
Work Step by Step
Let us assume the work function of the material is $\phi_{0}$
Given that for red light wavelength $\lambda_{red}= 700nm=700\times10^{-9}m=7\times10^{-7}m$
$E_{red}=hf_{red}=\frac{hc}{\lambda_{red}}$
putting $h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$,$\lambda_{red}=7\times10^{-7}m$
$E_{red}=hf_{red}=\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{7\times10^{-7}m}=2.8414\times10^{-19} J$
$1.6\times10^{-19} J$ is equal to $1eV$
$1J$ is equal to $\frac{1}{1.6\times10^{-19}}eV$
$2.8414\times10^{-19} J$ is equal to $\frac{2.8414\times10^{-19} }{1.6\times10^{-19}}eV=1.77589eV$
$E_{red}=hf_{red}=2.8414\times10^{-19}J=1.77589eV$
From equation 27.7
$hf= K_{max} + \phi_{0}$
$K_{max, red}=hf_{red}-\phi_{0} =1.77589eV-\phi_{0}$ equation number (1)
Given that for blue light wavelength $\lambda_{blue}= 400nm=400\times10^{-9}m=4\times10^{-7}m$
$E_{blue}=hf_{blue}=\frac{hc}{\lambda_{blue}}$
putting $h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$,$\lambda_{blue}=4\times10^{-7}m$
$E_{blue}=hf_{blue}=\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{4\times10^{-7}m}=4.9725\times10^{-19} J$
$1.6\times10^{-19} J$ is equal to $1eV$
$1J$ is equal to $\frac{1}{1.6\times10^{-19}}eV$
$4.9725\times10^{-19} J$ is equal to $\frac{4.9725\times10^{-19} }{1.6\times10^{-19}}eV=3.1078eV$
$E_{blue}=hf_{blue}=4.9725\times10^{-19}J=3.1078eV$
From equation 27.7
$hf= K_{max} + \phi_{0}$
$K_{max, blue}=hf_{blue}-\phi_{0} =3.1078eV-\phi_{0}$ equation number (2)
Given that $K_{max, blue}=2\times K_{max, red}$
putting the values from equation (1) and (2) we will get
$3.1078eV-\phi_{0}=(2\times 1.77589eV-\phi_{0})$
$3.1078eV-\phi_{0}=3.55178eV-2\phi_{0} $
$\phi_{0}=3.55178eV-3.1078eV=0.44398eV \approx0.44eV$
