AP Physics 2- 15.1 Quantum Theory and Wave-Particle Duality- FRQs- New Syllabus
Quantum Theory and Wave-Particle Duality AP Physics 2 FRQ
Unit 15: Modern Physics
Weightage : 15–18%
Question
| Trial | Frequency of Light | Metal Tested | \( \lambda_e \) \( (\times 10^{-10}\ \mathrm{m}) \) |
|---|---|---|---|
| \( 1 \) | \( f_A \) | Metal \( 1 \) | \( 6.9 \) |
| \( 2 \) | \( f_A \) | Metal \( 2 \) | \( 9.4 \) |
| \( 3 \) | \( f_B \) | Metal \( 1 \) | No electrons ejected |
| \( 4 \) | \( f_B \) | Metal \( 2 \) | No electrons ejected |
| \( 5 \) | \( f_C \) | Metal \( 1 \) | \( 5.3 \) |
| \( 6 \) | \( f_C \) | Metal \( 2 \) | \( 6.3 \) |
Most-appropriate topic codes (AP Physics 2):
• Topic \( 15.1 \) — Quantum Theory and Wave-Particle Duality (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \), through photon energy and de Broglie wavelength)
▶️ Answer/Explanation
(a)
The least frequency is \( \boxed{f_B} \), and the greatest frequency is \( \boxed{f_C} \).
No electrons are ejected for either metal when the incident light has frequency \( f_B \). Therefore, the photon energy \( hf_B \) must be below the threshold needed to eject electrons, so \( f_B \) is the smallest of the three frequencies.
For photoelectrons, a smaller de Broglie wavelength means a larger momentum, and therefore a larger kinetic energy. Comparing Trials \( 1 \) and \( 5 \) for Metal \( 1 \), the wavelength decreases from \( 6.9\times 10^{-10}\ \mathrm{m} \) to \( 5.3\times 10^{-10}\ \mathrm{m} \), so the electrons in Trial \( 5 \) have greater kinetic energy than in Trial \( 1 \). Since the metal is the same, the greater kinetic energy must come from the greater incident photon energy, so \( f_C > f_A \).
Combining these results gives \( f_B < f_A < f_C \).
(b)
Use the de Broglie relation
\( \lambda_e = \dfrac{h}{p} \)
so
\( p = \dfrac{h}{\lambda_e} \)
The kinetic energy can be written as
\( K = \dfrac{p^2}{2m_e} \)
Therefore,
\( K = \dfrac{1}{2m_e}\left(\dfrac{h}{\lambda_e}\right)^2 = \dfrac{h^2}{2m_e\lambda_e^2} \)
For Trial \( 1 \), \( \lambda_e = 6.9\times 10^{-10}\ \mathrm{m} \). Substituting \( h = 6.63\times 10^{-34}\ \mathrm{J\cdot s} \) and \( m_e = 9.11\times 10^{-31}\ \mathrm{kg} \):
\( K = \dfrac{(6.63\times 10^{-34})^2}{2(9.11\times 10^{-31})(6.9\times 10^{-10})^2} \)
\( K \approx 5.1\times 10^{-19}\ \mathrm{J} \)
So the maximum kinetic energy is
\( \boxed{5\times 10^{-19}\ \mathrm{J}} \)
which is about \( \boxed{3.2\ \mathrm{eV}} \).
(c)
The work function of Metal \( 1 \) is \( \boxed{\text{less than}} \) the work function of Metal \( 2 \).
For the same incident frequency, the electrons ejected from Metal \( 1 \) always have a smaller de Broglie wavelength than those ejected from Metal \( 2 \). A smaller wavelength means greater momentum and greater kinetic energy.
Since both metals receive photons of the same energy in a given pair of trials, the metal that gives the electrons the larger kinetic energy must have the smaller work function, because \( K_{\max} = hf – \phi \).
Therefore, Metal \( 1 \) has the smaller work function.