When photons of energy 4.25 eV strike the surface of a metal A , the ejected  photoelectrons  have maximum kinetic energy $T_A$ eV and de-Broglie wavelength ${\lambda }_A$  .  The maximum kinetic energy of photoelectrons liberated from  another metal B by  photons of energy 4.70eV is $T_B=\left(T_A-1.50eV\right)$ . If the de-Broglie wavelength of these  photoelectrons is${\lambda }_B=2{\lambda }_A$  , then,

In photoelectric effect, maximum kinetic energy of the ejected photoelectrons is given  by  $T=hv-\varphi$. 
 Thus,
  $T_A=4.25-{\varphi }_A$ ,     (1)  
$T_B=4.70-{\varphi }_B$   ,     (2)
 Given, 
 $T_B=T_A-1.50$  ,     (3)
$\frac{{\lambda }_A}{{\lambda }_B}=\frac{h/\sqrt{2m{T}_A}}{h/\sqrt{2m{T}_B}}=\sqrt{\frac{T_B}{T_A}}=\frac{1}{2}$   .   (4)
Solve the equations (3) and (4) to get $T_A=2eV$  and$T_B=0.5eV$  . Substitute $T_A$  in  equation (1) to get ${\varphi }_A=2.25eV$  and substitute $T_B$ in equation (2) to get ${\varphi }_B=4.2eV$ .