Photoelectric effect question

[![Photoelectric question](https://i.stack.imgur.com/7rNgw.png)](https://i.stack.imgur.com/7rNgw.png)


My attempt:- 1)At 0.68e15 Hz frequency of light,Metal A gives 7.2eV Kinetic Energy(KE).


2)At 1e15 Hz frequency of light, Metal B, gives 6eV Kinetic Energy.


3) At 1.1e15 Hz frequency of light, Metal C gives 5.5eV Kinetic Energy.


4)At1.2e15 Hz frequency of light, Metal D gives 5eV Kinetic energy.


Now $KE\_{max}=(\frac{hc}{\lambda}-\phi)$ where c= speed of light, $\lambda$=wavelength of light and $\phi$= work function of metal.


Now $7.2eV=(4.14e-15eVs \times 0.68e15 Hz- \phi)$ for metal A.The wavelength of light is 441 nm. Likewise we can compute wavelength of light incident on Metal B, C and D.


But we don't know the work functions of Metal A,B, C and D.


So, how to answer this question?


