Quantum Theory and Wave-Particle Duality AP Physics 2 MCQ

Quantum Theory and Wave-Particle Duality AP Physics 2 MCQ – Exam Style Questions etc.

Quantum Theory and Wave-Particle Duality AP Physics 2 MCQ

Unit 15: Modern Physics

Weightage : 15–18%

AP Physics 2 Exam Style Questions – All Topics

Question

An experiment is set up such that a beam of monochromatic light passes through a diffraction grating with 3000 lines/cm, creating a diffraction pattern on a screen located 0.1 meter from the grating. The distance between the central maximum and the next nearest bright line on the screen is 1.32 cm.

(a) Determine the wavelength of the incident light.

(b) What is the energy of each incident photon (in eV)?

The incident light is created by a filtered mercury arc lamp. A few energy levels of mercury are shown at right.

(c) Which electron transition in the lamp is most likely responsible for creation of the photons striking the diffraction grating?

(d) The diffraction grating is replaced with a new grating containing 2000 lines/cm. Does the distance between maxima
              ___ increase
              ___ decrease
              ___ remain the same
Explain your reasoning.

Answer/Explanation

Ans:
(a) First find the spacing between slits:\(\frac{ .01m}{ 3000lines} = 3.33×10^{−6} m\)

Next determine a value for sin θ given your geometric setup:
\(θ = tan^{−1}\frac{ 0.0132m}{0.1m}= 7.52°→ sinθ = sin(7.52°) = 0.1308\)

Then solve for the wavelength of light using the diffraction grating equation:

(b) \(E = hf = \farc{hc}{λ} =\frac {1240eV . nm}{436nm} = 2.84eV\)

(d) Decrease. With fewer slits per cm, the width of the slits increases. Therefore, the angle must decrease.

Question

An X-ray photon with frequency f undergoes an elastic collision with a stationary electron (mass \(m_e\) ) and is scattered. The frequency of the scattered X-ray photon is f ’.

(a) Determine the kinetic energy of the electron after the collision in terms of f, f ’, and fundamental constants.

(b) Determine the magnitude of the momentum of the scattered electron in terms of f, f ’, \(m_e\) , and fundamental constants.

Answer/Explanation

Ans:
(a) Utilizing conservation of energy: \(hf = K_{electron} + h{f}’→ K_{electron} = hf −h{f}’= h( f − {f}’)\)

(b) First find the speed of the electron: \(K =\frac{ 1}{ 2} mv^2 → v = \sqrt{\frac{2K}{ m_e}} = \sqrt{\frac{2h( f − {f}’)}{ m_e}}\)

Next solve for the momentum: \(p = mv = m_e\sqrt{\frac{2h( f − {f}’)}{ m_e}}=\sqrt{2m_eh( f − {f}’)}\)

Quantum Theory and Wave-Particle Duality AP Physics 2 MCQ