The graph detween 1/ $\lambda$, and stopping potential (V) of three metals having work functions f₁, f₂ and f₃ in an experiment of photo-electric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct?

According to Einstein’s photoelectric equation

$\frac{hc}{\lambda }-\cancel{o}=eVwhere\cancel{o}=workfunction$

$Vand\left(\frac{1}{\lambda }\right)$ relation represents a straight line

Slope of line$=\frac{hc}{e}or\tan \theta =\frac{hc}{e}$

At $V_0=0, \lambda ={\lambda }_0,\varphi =\frac{hc}{\lambda }$

${\varphi }_1:{\varphi }_2:{\varphi }_3=\frac{hc}{{\lambda }_{01}}:\frac{hc}{{\lambda }_{02}}:\frac{hc}{{\lambda }_{03}}=0.001:0.002:0.004=1:2:4$

$\frac{1}{{\lambda }_{01}}=0.001n{m}^{-1}\Rightarrow {\lambda }_{01}={10}^{-6}m$

$\frac{1}{{\lambda }_{02}}=0.002n{m}^{-1}\Rightarrow {\lambda }_{02}=0.5\times {10}^{-6}m$

$\frac{1}{{\lambda }_{03}}=0.004n{m}^{-1}\Rightarrow {\lambda }_{03}=0.25\times {10}^{-6}m$

Now ${\lambda }_{violet}=0.4x{10}^{-6}m$

For ejection of photoelectrons,

${\lambda }_{violet}<{\lambda }_0$

So, violet colour can eject photoelectrons from metals 1, 2 but not from metal 3