Given,
Energy of two photons are E1=4eV and E2=2.5eV
The ratio of maximum speeds of the photoelectrons emitted in the two cases is v2v1=2
Using Einstein equation of photoelectric effect,
KEmax=21mv2=E−ϕ...(1)
Where, ϕ is the work function of metal and E is the energy of photon
Now using equation for both the cases we get,
21mv12=4−ϕ…(2)
21mv22=2.5−ϕ...(3)
Dividing equation (2) and (3) and substitute given values, we get,
v22v12=2.5−ϕ4−ϕ=(2)2
⇒3ϕ=6
⇒ϕ=2eV