Question: (10 points, suggested time 20 minutes)
Light and matter can be modeled as waves or as particles. Some phenomena can be explained using the wave model, and others can be explained using the particle model.
(a) Calculate the speed, in m/s, of an electron that has a wavelength of 5.0 nm.
(b) The electron is moving with the speed calculated in part (a) when it collides with a positron that is at rest. A positron is a particle identical to an electron except that its charge is positive. The two particles annihilate each other, producing photons. Calculate the total energy of the photons.
(c) A photon approaches an electron at rest, as shown above on the left, and collides elastically with the electron. After the collision, the electron moves toward the top of the page and to the right, as shown above on the right, at a known speed and angle. In a coherent, paragraph-length response, indicate a possible direction for the photon that exists after the collision and its frequency compared to that of the original photon. Describe the application of physics principles that can be used to determine the direction of motion and frequency of the photon that exists after the collision.
▶️Answer/Explanation
Ans:
\(\lambda =\frac{h}{p}\) 5 × 10-9 m = \(\frac{6.63\times 10^{-34}J.5}{(9.11\times 10^{-31}kg)V}\) p = mv
4.555× 10-39 V = 6.63 × 10-34
V = 14554.3 m/s
(b)
E = Δmc2 Δm = 2 (9.11 × 10-31 ) = 1.822 × 10-34
Po = Pf
14554 (me) = mv
E= 1.822 × 10-30 (3×108)2
E = 1.64 × 10-13 J
(c)
The initial momentum has to equal to final momentum in all directions. The frequency of the original will be greater than the final frequency as energy must be conserved and it depends on frequency. The photon will travel back towards the bottom of the page as to conserve momentum of the vertical axis but the angle will be less than the election to conserve horizontal momentum.
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) \(E = hf = \farc{hc}{λ} =\frac {1240eV . nm}{436nm} = 2.84eV\)
(c) Level f to level c: \(E_{photon} = E_i − E_f =−2.68eV −(−5.52eV) = 2.84eV\)
(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.
(c) Determine the de Broglie wavelength of the 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: \(p = mv = m_e\sqrt{\frac{2h( f − {f}’)}{ m_e}}=\sqrt{2m_eh( f − {f}’)}\)
(c) \(λ = \frac{h}{p}=\frac{ h}{\sqrt{2m_eh( f − {f}’)}}\)