The Bohr Model of Atomic Structure AP Physics 2 MCQ – Exam Style Questions etc.
The Bohr Model of Atomic Structure AP Physics 2 MCQ
Unit 15: Modern Physics
Weightage : 15–18%
Exam Style Practice Questions ,The Bohr Model of Atomic Structure AP Physics 2 MCQ
Question
A sodium photoelectric surface with work function 2.3 eV is illuminated by electromagnetic radiation and emits electrons. The electrons travel toward a negatively charged cathode and complete the circuit shown above. The potential difference supplied by the power supply is increased, and when it reaches 4.5 V, no electrons reach the cathode.
a. For the electrons emitted from the sodium surface, calculate the following.
i. The maximum kinetic energy
ii. The speed at this maximum kinetic energy
b. Calculate the wavelength of the radiation that is incident on the sodium surface.
c. Calculate the minimum frequency of light that will cause photoemission from this sodium surface.
Answer/Explanation
Ans:
a) i. K = Vstop q … K = (4.5V)(1e) = 4.5 eV
ii. K = ½ mv2 … 4.5eV*(1.6×10–19 J/eV) = ½ (9.11×10–31)v2 … v = 1.26×106 m/s
b) From K = Ein – Φ … 4.5 eV = Ein – 2.3 eV … Ein = 6.8 eV
E = hc / λ … 6.8 =1240/ λ … λ = 182 nm
c) Emin = Φ … hfo = Φ … (4.14×10–15) fo = 2.3 … fo = 5.56×1014 Hz.
Question
A transmission diffraction grating with 600 lines/mm is used to study the line spectrum of the light produced by a hydrogen discharge tube with the setup shown above. The grating is 1.0 m from the source (a hole at the center of the meter stick). An observer sees the first-order red line at a distance yr = 428 mm from the hole.
a. Calculate the wavelength of the red line in the hydrogen spectrum.
b. According to the Bohr model, the energy levels of the hydrogen atom are given by En = – 13.6 eV/n2, where n is an integer labeling the levels. The red line is a transition to a final level with n = 2. Use the Bohr model to determine the value of n for the initial level of the transition.
c. Qualitatively describe how the location of the first-order red line would change if a diffraction grating with 800 lines/mm were used instead of one with 600 lines/mm.
Answer/Explanation
Ans:
a) Diffraction grating: Since the screen distance is 1 m and the first order line is at 0.428 m, the angle is not small and the small angle approximation cannot be used. Instead we find the angle with tan θ and use mλ = d sin θ. First find d. d = 600 lines / mm = 1/600 mm / line = 0.00167 mm / line = 1.67×10–6 m / line.
tan θ = o/a … tan θ = 0.428 / 1 … θ = 23° … Then, mλ = d sin θ … (1) λ = (1.67×10–6) sin 23°
λ = 6.57×10–7 m = 657 nm
b) First find the energy of the red light emission. E = hc / λ … E = 1240/657 = 1.89 eV emitted In the problem, the n=2 level energy state is given by En = -13.6 /n2 = E2 = –13.6 / 22 = –3.4 eV
Since the red light was an emission of 1.89 eV, this is the energy that was removed from the initial level to end up in the –3.4 eV level, → it must have originated in a higher energy level (less negative) which can be found by adding back that emitted energy –3.4 + 1.89 = – 1.51. Plugging this back into the provided equation, En = -13.6 /n2 –1.51 = –13.6 /n2 … n2 = 9, so n= 3 is the original level.
c) Referring to the calculation of d from part a .. d = 1/800 mm / line which is a smaller d value. Less d means sin θ must increase so the angle is larger and the location of the line would be further out