In a historical experiment to determine Planck's constant, a metal...

K{E}_\max = \dfrac{hC}{\lambda} - ϕ

$e\left({V}_{0}\right) = \dfrac{hC}{\lambda} - ϕ$
$1.6 \times {10}^{- 19} \times 2 = \dfrac{h \times 3 \times {10}^{8}}{3000 \times {10}^{- 10}} - ϕ$ .....(i)
$1.6 \times {10}^{- 19} \times 1 = \dfrac{h \times 3 \times {10}^{8}}{4000 \times {10}^{- 10}} - ϕ$ .......(ii)
From (ii) $ϕ = \dfrac{h \times 3 \times {10}^{8}}{4000 \times {10}^{- 10}} - 1.6 \times {10}^{- 19}$

Using value of $ϕ$ in equation (i), we get
$1.6 \times {10}^{- 19} \times 2 = \dfrac{h \times 3 \times {10}^{8}}{3000 \times {10}^{- 10}} - \dfrac{h \times 3 \times {10}^{8}}{4000 \times {10}^{- 10}} + 1.6 \times {10}^{- 19}$
$1.6 \times {10}^{- 19} = \dfrac{h \times 3 \times {10}^{8}}{{10}^{- 7}}\left(\dfrac{1}{3} - \dfrac{1}{4}\right) = \dfrac{h \times 3 \times {10}^{8}}{{10}^{- 7}}\left[\dfrac{4 - 3}{12}\right]$
$1.6 \times {10}^{- 19} = \dfrac{h \times 3 \times {10}^{8}}{{10}^{- 7}} \times \dfrac{1}{12}$
$1.6 \times 4 \times \dfrac{{10}^{- 19} \times {10}^{- 7}}{{10}^{8}} = h$
$6.4 \times {10}^{- 34}Js = h$