Photons of energy  2eV fall on a metal plate and release photo electrons with a maximum velocity v, by decreasing the wave length by 25% the maximum velocity of photo electrons is doubled. The work function of metal of the material of plate in eV  is

$v_1=v,v_2=2v$

We know that

$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-w \left[here w=work function\right]$

$\begin{array}{l}\therefore \frac{v_1^2}{v_2^2}=\frac{\frac{12400}{{\lambda }_1}-w}{\frac{12400}{{\lambda }_2}-w}\\ Given : energy of photon=\frac{12400}{{\lambda }_1}=2eV\\ New wavelength={\lambda }_2=\frac{3}{4}{\lambda }_1\\ New energy of photon=\frac{12400}{{\lambda }_2}=\frac{12400}{\frac{3}{4}{\lambda }_1}=\frac{4}{3}\left(2\right)eV\\ \Rightarrow \frac{v^2}{4v^2}=\frac{2-w}{\frac{4}{3}\left(2\right)-w}\\ \Rightarrow 8-4w=\frac{8}{3}-w\\ \Rightarrow 8-\frac{8}{3}=4w-w\\ \Rightarrow \frac{16}{3}=3w\\ \Rightarrow w=\frac{16}{9}=1.78eV\end{array}$