Question

The green graph and the purple graph pictured below represent the photoelectric effect for two different metals (G and P).
If electromagnetic radiation of frequency $11\times {10}^{14}Hz$ strikes both metal surfaces, which of the following statements must be true?

• A. The radiation will have to be more intense to knock electrons off the surface of metal P than metal G
• B. No electrons will be knocked off the surface of either metal as a result of the radiation
• C. Electrons will be knocked off both metal surfaces, but they will be knocked off at a greater rate from the surface of metal G
• D. Electrons will be knocked off the surface of metal G by the radiation, but will not be knocked off the surface of metal P
• E. Electrons will be knocked off both metal surfaces by the radiation, but the most energetic electrons knocked off the surface of metal G will have more kinetic energy than the most energetic electrons knocked off the surface of metal P

Answer 1

The correct option is E Electrons will be knocked off both metal surfaces by the radiation, but the most energetic electrons knocked off the surface of metal G will have more kinetic energy than the most energetic electrons knocked off the surface of metal P

As the frequency of the EM radiation used is greater than the threshold frequency of both the metal P and G, thus electrons will be knocked off both metal surfaces by the radiation.

Maximum kinetic energy of the photoelectrons $K.E_{max}=h\nu -\varphi$

Also from graph, we can state that work function $\left(\varphi \right)$ of metal P is greater than that of G i.e ${\varphi }_P>{\varphi }_G$

$⟹$ $K.E_G{}_{max}>K.E_{P_{max}}$

Hence option E is correct.