Q.Two metals $P$ and $Q$ have work functions such that $\phi_P > \phi_Q$. Which metal has the higher threshold frequency? If the frequency of incident light is increased gradually from zero, which of the two metals begins photoelectric emission first? Explain your reasoning.

Threshold frequency and work function. From Einstein's equation, threshold frequency is $\nu_0=\phi_0/h$ -- since $h$ is the same universal constant for every metal, $\nu_0$ is directly PROPORTIONAL to $\phi_0$. A metal with a LARGER work function therefore has a correspondingly LARGER threshold frequency.

Applying this to $P$ and $Q$. Given $\phi_P>\phi_Q$, it follows directly that $\nu_{0,P}>\nu_{0,Q}$ -- metal $P$ has the higher threshold frequency. …

Use $\nu_0=\phi_0/h$ to relate the two metals' threshold frequencies to their given work functions, then reason about which thre …

• Reasoning that the HIGHER-work-function metal ($P$) should emit first, by mistakenly associating 'higher $\phi_0$' with 'easier to trigger' -- it is the opposite: a HIGHER work function means a HIGHER (harder to reach) threshold frequency. …