Physics Problems & Solutions: #17

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Quantum Mechanics - Probability

The wave function for a particle constrained to move in one dimension is shown in the graph (Ψ = 0 for x ≤ 0 and x ≥ 5). What is the probability that the particle would be found between x = 2 and x = 4?

A. 17/64
B. 25/64
C. 5/8
D. v(5/8)
E. 13/16

(GR9677 #17)

Solution:

Probability, P ∼ ψ²

Probability to find the particle between x = 2 and x = 4 (unnormalized probability):
ψ²= 2² + 3²= 13

Total probability (normalized probability):
ψ²= 1² + 1² + 2² + 3² + 1²= 16

P = unnormalized probability / normalized probability = 13/16

Answer: E

Lab Methods - Oscilloscope

The outputs of two electrical oscillators are compared on an oscilloscope screen. The oscilloscope spot is initially at the center of the screen. Oscillator Y is connected to the vertical terminals of the oscilloscope and oscillator X to the horizontal terminals. Which of the following patterns could appear on the oscilloscope screen, if the frequency of oscillator Y is twice that of oscillator X?

(GR9277 #17)

Solution:

→

Answer:

Nuclear & Particle Physics - Radioactivity

Suppose that

decays by natural radioactivity in two stages to $_{Z-1}^{A-4}Y$ . The two stages would most likely be which of the following?

	First Stage	Second Stage
A.	β⁻ emission with an antineutrino	α emission
B.	β⁻ emission	α emission with a neutrino
C.	β⁻ emission	γ emission
D.	Emission of a deuteron	Emission of two neutrons
E.	α emission	γ emission

(GR9677 #30)

Solution:

→ First Stage → Second Stage → $_{Z-1}^{A-4}Y$

by natural decay.

A. TRUE.
β⁻ decay: emission of an electron and electron anti-neutrino,

$_Z^AX \rightarrow {_{Z+1}^A}X'+{_{-1}^0}\beta+\bar{\nu}$

α decay: emission of a Helium nucleus resulting in the loss of two protons and two neutrons,

$_Z^AX \rightarrow {_{Z-2}^{A-4}}X'+{_{2}^4}He$

Therefore,

$\\_Z^AX \rightarrow \beta^-$ decay$\rightarrow \alpha $ decay$\rightarrow {_{Z-1}^{A-4}}Y\\\\_Z^AX \rightarrow {_{Z+1}^A}X'\rightarrow {_{Z+1-2}^{A-4}}X''\rightarrow {_{Z-1}^{A-4}}Y$

B. FALSE.
β⁻ decay always emits an anti-neutrino
α decay does not emit neutrino

C. FALSE.
γ decay: photon emission with no change in mass number A or atomic number Z

$_Z^AX \rightarrow{ _Z^AX}'+\gamma$

Therefore,

$_Z^AX \rightarrow {_{Z+1}^A}X'\rightarrow {_{Z+1}^A}X''\rightarrow {_{Z+1}^A}Y$

D. FALSE.
Deuteron decay:
Deuteron (nucleus of deuterium) is a stable particle.
Deuteron decay is rare, not natural.

E. FALSE.

$_Z^AX \rightarrow {_{Z-2}^{A-4}}X'\rightarrow {_{Z-2}^{A-4}}X''\rightarrow {_{Z-2}^{A-4}}Y$

Answer: A

Nuclear & Particle Physics - Electron Configuration

The ground state electron configuration for phosphorus, which has 15 electrons, is

A. 1s² 2s² 2p⁶ 3s¹ 3p⁴
B. 1s² 2s² 2p⁶ 3s² 3p³
C. 1s² 2s² 2p⁶ 3s² 3d³
D. 1s² 2s² 2p⁶3s¹ 3d⁴
E. 1s² 2s² 2p⁶ 3p²3d³

(GR0177 #17)

Solution:

Phosphorus (15 electrons): 1s² 2s² 2p⁶3s² 3p³

Answer: B

Note:

Electron configuration

	l = 0	l = 1	l = 2	l = 3	# of electrons
n = 1	1s²				2
n = 2	2s²	2p⁶			8
n = 3	3s²	3p⁶	3d¹⁰		18
n = 4	4s²	4p⁶	4d¹⁰	4f¹⁴	32