IBDP Physics- E.2 Quantum physics- IB Style Questions For HL Paper 1A -FA 2025

Correct Answer: \( \boxed{\mathrm{D}} \)

The photoelectric equation is

\(E_{\max}=hf-\phi\)

where \(h\) is Planck’s constant, \(f\) is the frequency of the incident light and \(\phi\) is the work function of the metal.

The threshold frequency is obtained when \(E_{\max}=0\):

\(f_0=\dfrac{\phi}{h}\)

Since surface H requires a smaller minimum photon energy, its work function is smaller:

\(\phi_H<\phi_L\)

Therefore,

\(f_{0,H}<f_{0,L}\)

So the line for H must cross the frequency axis to the left of the line for L.

Also, from \(E_{\max}=hf-\phi\), the gradient of an \(E_{\max}\)-against-\(f\) graph is \(h\), which is the same for both metals.

Therefore, the two lines must have the same gradient, with H having the lower threshold frequency.

Hence, the correct answer is \( \boxed{\mathrm{D}} \).