In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of…

$K{E}_{max}=\frac{hC}{\lambda }-\varphi$

$e\left(V_0\right)=\frac{hC}{\lambda }-\varphi$
$1.6\times {10}^{-19}\times 2=\frac{h\times 3\times {10}^8}{3000\times {10}^{-10}}-\varphi$ .....(i)
$1.6\times {10}^{-19}\times 1=\frac{h\times 3\times {10}^8}{4000\times {10}^{-10}}-\varphi$ .......(ii)
From (ii) $\varphi =\frac{h\times 3\times {10}^8}{4000\times {10}^{-10}}-1.6\times {10}^{-19}$

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