Radiation of wavelength lambda is incident on a photocell. The fastest emitted electron has speed υ.

Explanation:

The kinetic energy can be written as 

E₁=hc/λ-ϕ  where λ is the wavelength of radiation ϕ is the working function of the photocell.

⇒1/2 mu²=hc/λ-ϕ

Now the wavelength 3λ/4.

So the kinetic energy 

E₂=hc/(3λ/4)-ϕ

⇒E₂=4hc/3λ-ϕ

⇒1/2 mu₂²=4hc/3λ-ϕ

E₂>4/3 E₁ so u₂>u(4/3)^1/2