IBDP Physics- E.2 Quantum physics- IB Style Questions For HL Paper 1A -FA 2025
Question
(B) \( \sqrt{\frac{E – \Phi}{m_e}} \)
(C) \( \sqrt{\frac{E – \Phi}{m_e h}} \)
(D) \( \sqrt{\frac{2(E – \Phi)}{m_e h}} \)
▶️ Answer/Explanation
Photoelectric equation: \( E = \Phi + KE_{\text{max}} \)
\( KE_{\text{max}} = E – \Phi = \frac{1}{2}m_ev^2 \)
\( v = \sqrt{\frac{2(E – \Phi)}{m_e}} \)
✅ Answer: (A)
Question
| Option | Maximum kinetic energy | Work function |
|---|---|---|
| A | less than \(2E_{\max}\) | unchanged |
| B | less than \(2E_{\max}\) | greater than \(W\) |
| C | greater than \(2E_{\max}\) | unchanged |
| D | greater than \(2E_{\max}\) | greater than \(W\) |
▶️ Answer / Explanation
The photoelectric equation is:
\[ E_{\max} = hf – W \]
When the frequency is doubled:
\[ E’_{\max} = 2hf – W \]
Compare with: \[ 2E_{\max} = 2(hf – W) = 2hf – 2W \]
Since \(E’_{\max} = 2hf – W > 2hf – 2W\), the new maximum kinetic energy is greater than \(2E_{\max}\).
The work function \(W\) is a property of the metal and does not depend on the frequency of the incident light, so it remains unchanged.
✅ Answer: C
Question
(B) It generates an alternating current.
(C) It absorbs energy over a range of photon frequencies.
(D) It can be used to store energy in a secondary cell.
▶️ Answer / Explanation
A photovoltaic (solar) cell works by the photoelectric effect, converting light energy directly into electrical energy.
• The output power depends on the surface area of the cell ✔️
• It absorbs photons over a range of frequencies ✔️
• The electrical energy produced can be stored using a secondary (rechargeable) cell ✔️
However, a photovoltaic cell produces direct current (DC), not alternating current. AC output requires an external inverter.
✅ Answer: (B)
