when a surface 1cm thick is illuminated with light of wavelength lambda, the stopping potential is Vo. When the surface is illuminated with light of wavelength 3lambda, the stopping potential is Vo/6. find threshold wavelength for the metallic surface

I'm not getting  the answer could you please show me how you did it? 

hi
we can write 2 equations from given data
Vo=hc/Lo - hc/L --1 
and 
Vo/6 = hc/Lo - hc/3L  --2
substitute value of Vo in 2 as 
hc/6Lo -hc/6L = hc/Lo -hc/3L
therefore 
5hc/6Lo = hc/6L
thereore Lo = 5L
(L=lambda)
 

Use Einstein's equation.......
V=(hc/e)(1/$ - 1/@)..........   Im using $ for incident wavelength and @ for cutoff wavelength

In case first:
V₀=(hc/e) (1/$ - 1/@)............. (1)
In second case:
V₀/6= (hc/e) (1/3$ - 1/@)........... (2)

Divide eq 1 by 2..... V₀ and hc/e would cancel out.....
You'd get eqution with $ and @....... 
Solving you'd get @=5$
 

Thank you!