When radiation of wavelength λ is incident on a metallic surface,...

{V}_{S} = hv - ϕ$</p><p>$4 . 8 = \dfrac{hc}{\lambda} - ϕ . . . ( i )$</p><p>$1 . 6 = \dfrac{hc}{2\lambda} - ϕ . . . ( ii )$</p><p>Using above equation$( i ) - ( ii )$</p><p>$3 . 2 = \dfrac{hc}{\lambda} - \dfrac{hc}{2\lambda}$</p><p>$3 . 2 = \dfrac{hc}{2\lambda} . . . ( iii )$</p><p>\left[\lambda = \dfrac{hc}{6 . 4}\right]$

Put in equation $( ii )$

$ϕ = 1 . 6$

$\dfrac{hc}{{\lambda}_{th}} = 1 . 6$

${\lambda}_{th} = \dfrac{hc}{1 . 6}$

$= \left(\dfrac{hc}{6 . 4}\right) \times 4 = 4\lambda$