Physics Problems & Solutions: #18

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Quantum Mechanics - Potential Wall

Consider a potential of the form

V(x) = 0, x ≤ a
V(x) = V_0,a < x < b
V(x) = 0, x ≥ b

As shown in the figure above. Which of the following wave functions is possible for a particle incident from the left with energy E < V₀.

(GR9677 #18)

Solution:

A. Classic not QM potential → FALSE
B. No decrease in amplitude → FALSE
C. Decay exponentially inside the wall, decrease amplitude (fits E < V₀) → TRUE

D. QM Oscillator harmonics → FALSE
E. Cosine wave function not QM potential → FALSE

Answer: C

Electromagnetism - Impedance

In transmitting high frequency signals on a coaxial cable, it is important that the cable be terminated at an end with its characteristic impedance in order to avoid

A. Leakage of the signal out of the cable
B. Overheating of the cable
C. Reflection of signals from the terminated end of the cable
D. Attenuation of the signal propagating in the cable
E. Production of image currents in the outer conductors

(GR9277 #18)

Solution:

Characteristic impedance of a transmission line is the ratio of the voltage and current of a wave travelling along the line.

When the wave reaches the end of the line, a reflected wave could travels back in the opposite direction. It could develop an interference and cause the voltage to fluctuates.

To avoid this problem, the receiving end of a coaxial cable should be terminated using a resistance value equals to its characteristic impedance.

Answer: C

Quantum Mechanics - Schrodinger Equation

The wave function ψ(x) = A exp (−b²x²/2), where A and b are real constants, is a normalized eigenfunction of the Schrodinger equation for a particle of mass M and energy E in a one dimensional potential V(x) such that V(x) = 0 at x = 0. Which of the following is correct?

A. V = ħ²b⁴/2M
B. V = ħ²b⁴x²/2M
C. V = ħ²b⁶x⁴/2M
D. E = ħ²b²(1 − b²x²)
E. E = ħ²b⁴/2M

(GR8677 #18)

Solution:

Schrodinger Equation: $-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+V(x)\psi(x)=E\psi(x)$

$\begin{aligned} \frac{\partial\psi}{\partial x}&=\frac{\partial}{\partial x}e^{-b^2x^2/2}=-b^2x\psi \\\frac{\partial^{2}\psi}{\partial x^{2}}&=-b^2\frac{\partial}{\partial x}x\psi=-b^2\left( x\frac{\partial\psi}{\partial x}+\psi\frac{\partial x}{\partial x}\right ) \\\frac{\partial^{2}\psi}{\partial x^{2}}&=-b^2(-b^2x^2\psi+\psi)=(b^4x^2-b^2)\psi\end{aligned}$

Schrodinger Equation: $-\frac{\hbar^{2}}{2m}(b^4x^2-b^2)\psi(x)+V(x)\psi(x)=E\psi(x)$

To find E → V(x = 0) = 0:

$\\-\frac{\hbar^{2}}{2m}(0-b^2)+0=E \\\rightarrow E=\frac{\hbar^{2}b^2}{2m}$

To find V(x):
$\\-\frac{\hbar^{2}}{2m}(b^4x^2-b^2)+V(x)=E \\\\-\frac{\hbar^{2}b^4x^2}{2m}+\frac{\hbar^{2}b^2}{2m}+V(x)=E \\\\-\frac{\hbar^{2}b^4x^2}{2m}+E+V(x)=E \\\\\rightarrow V(x)=\frac{\hbar^{2}b^4x^2}{2m}$

Answer: B

Nuclear & Particle Physics - Helium

The energy required to remove both electrons from the Helium atom in its ground state is 79.0 eV. How much energy is required to ionize Helium (i.e. to remove one electron)?

A. 24.6 eV
B. 39.5 eV
C. 51.8 eV
D. 54.4 eV
E. 65.4 eV

(GR0177 #18)

Solution:

Helium: 2 Protons, 2 Neutrons, 2 Electrons.

The energy required to remove one electron from He in its ground state, leaving behind He⁺ (a Hydrogen like atom)

E_n = 13.6 Z²/n² eV

Z = Helium atomic number = 2
n = 1 (ground state)

E₁ = 13.6(4) eV = 54.4 eV

The energy required to remove both electrons from He in its ground state leaving behind He⁺⁺ ion = 79 eV.

Thus, the energy required to remove one electron: 79 − 54.4 = 24.6 eV

Answer: A