Question

Take h to have an exact value of 6.63×10−34J⋅s for significant figure purposes, and use hc=1.24×103eV⋅nm (three significant figures). When a certain photoelectric material is illuminated with red light (λ=715) and then blue light (λ=405), it is found that the maximum kinetic energy of the photoelectrons resulting from the blue light is twice that from red light. ϕ0 = eV

Answer 1

he photoelectric equation is
1/2(mv^2)=h*c/lamda-phi

where phi the work function of the material in joules

eq 1 for red light

1/2(mv^2)=h*c/715x10^-9 - phi

eq 2 for blue light

2(1/2(mv^2)=h*c/405x10^-9 - phi

if we divide the two equations we find

1/2=(h*c/715x10^-9 - phi )/(h*c/405x10^-9 - phi )

=> h*c/405x10^-9 - phi =2*(h*c/715x10^-9 - phi )

=> h*c/405x10^-9 - phi =2*h*c/715x10^-9 - 2*phi

=> h*c/405x10^-9 - 2*h*c/715x10^-9= -2phi +phi

=> phi = 2*h*c/715x10^-9 - h*c/405x10^-9

phi = h*c( 2/715x10^-9 - 1/405x10^-9) = 1.24x10^3x10^-9x10^9x(2/715 - 1/405)

phi = 0.406803 eV