////Given
lembda=3500*10**-10                               //m
h=6.6*10**-34
c=3*10**8                                         //m/s

//calculation 
E=((h*c)/lembda)/1.6*10**-19

//Result
printf("\n Energy is  %0.2f ev",E*10**38)
printf("\n 1.9 ev < E < 4.2 ev,only metal B will yield photoelectrons")