When 1 cm thick surface is illuminated with light of wavelength λ, the stopping potential is V. When the same surface is illuminated by light of wavelength 2λ, the stopping potential is V/3. Threshold wavelength for metallic surface is:

4λ

According to the question,

$eV=hc(\frac{1}{λ}-\frac{1}{λ_0})$  (i)

and $\frac{eV}{3}=hc(\frac{1}{2λ}-\frac{1}{λ_0})$  (ii)

Dividing eq. (i) by eq. (ii), we get

$3=\frac{(\frac{1}{λ}-\frac{1}{λ_0})}{(\frac{1}{2λ}-\frac{1}{λ_0})}$

or $3(\frac{1}{2λ}-\frac{1}{λ_0})=\frac{1}{2λ}-\frac{1}{λ_0}$

or $\frac{3}{2λ}-\frac{1}{λ}=\frac{3}{λ_0}-\frac{1}{λ_0}$

or $\frac{1}{2λ}=\frac{2}{λ_0}$

Threshold wavelength for metallic surface

$λ_0=4λ$