$$K_1$$ and $$K_2$$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $$\lambda_1$$ and $$\lambda_2$$, respectively. If $$\lambda_1 = 2\lambda_2$$ then the work function of material is given by :

Use Einstein’s photoelectric equation:

K_{\max}=\frac{hc}{\lambda}$$

For the two դեպ:

$$K_1=\frac{hc}{\lambda_1}-\phi,\quad $$

$$K_2=\frac{hc}{\lambda_2}$$

Given:

$$\lambda_1=2\lambda_2$$

So,

$$\frac{hc}{\lambda_1}=\frac{hc}{2\lambda_2}=\frac{1}{2}\frac{hc}{\lambda_2}$$

Now rewrite:

$$K_1=\frac{1}{2}\frac{hc}{\lambda_2}$$

$$K_2=\frac{hc}{\lambda_2}$$ 

Let:

$$X=\frac{hc}{\lambda_2}$$

Then:

$$K_1=\frac{X}{2}-\phi,\quad K_2=X-\phi$$

$$K_2-K_1=(X-\phi)-\left(\frac{X}{2}-\phi\right)=\frac{X}{2}$$

$$\Rightarrow X=2(K_2-K_1)$$

$$\phi=X-K_2$$

Substitute:

$$\phi=2(K_2-K_1)-K_2$$