Physics Problems & Solutions: Energy

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Electromagnetism - Capacitor

The capacitor in the circuit shown above is initially charged. After closing the switch, how much time elapses until one-half of the capacitor’s initial stored energy is dissipated?

A. RC
B. RC/2
C. RC/4
D. RC ln 2
E. ½ RC ln 2

(GR9277 #11)

Solution:

U_t= ½U₀ , t = ?
Energy of capacitor: U = ½CV²

U_t = ½U₀→ ½CV_t² = ½ × ½CV₀2
V_t² = ½ V₀2

Potential difference of capacitor: V_t= V₀e^−t/RC
V_t²= V₀2e^−2t/RC
½ V₀2= V₀2e^−2t/RC→ ½ = e^−2t/RC
ln ½ = − 2t/RC → ln 2 = 2t/RC
t = ½ RC ln 2

Answer: E

Classical Mechanics - Lagrangian

A particle of mass m on the Earth’s surface is confined to move on the parabolic curve y = ax², where y is up. Which of the following is a Lagrangian for the particle?

A. $L = \frac{1}{2} m\dot{y} ^2 \left (1+\frac{1}{4ay} \right )-mgy$

B. $L = \frac{1}{2} m\dot{y} ^2 \left (1-\frac{1}{4ay} \right )-mgy$

C. $L = \frac{1}{2} m\dot{x} ^2 \left (1+\frac{1}{4ay} \right )-mgx$

D. $L = \frac{1}{2} m\dot{x} ^2 \left (1+4a^2x^2 \right )+mgx$

E. $L = \frac{1}{2} m\dot{x} ^2 +\frac{1}{2} m\dot{y} ^2+mgy$

(GR9277 #44)

Solution:

Kinetic Energy, $T = \frac{1}{2} m(\dot{x} ^2 +\dot{y} ^2)$

Potential Energy,

Lagrangian, $L=T-U=\frac{1}{2} m(\dot{x} ^2 +\dot{y} ^2)-mgy$

Given the curve y = ax²,

$\begin{aligned} \dot{y}&=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=2ax\dot{x}\\\rightarrow \dot{x}&=\frac{\dot{y}}{2ax} \\\dot{x}^2&=\frac{\dot{y}^2}{4a^2x^2}=\frac{\dot{y}^2}{4ay} \end{aligned}$

$\begin{aligned} L&=\frac{1}{2} m\left (\frac{\dot{y} ^2}{4ay}+\dot{y}^2 \right ) -mgy \\&=\frac{1}{2} m\dot{y} ^2\left (\frac{1}{4ay}+1 \right ) -mgy \end{aligned}$

Answer: A

Classical Mechanics - Energy

A ball is dropped from a height h. As it bounces off the floor, its speed is 80% of what it was just before it hit the floor. The ball will then rise to a height of most nearly

A. 0.94 h
B. 0.80 h
C. 0.75 h
D. 0.64 h
E. 0.50 h

(GR9277 #45)

Solution:

Conservation of energy before and when the ball hits the ground:

$\begin{aligned} \frac{1}{2}m \cdot 0+mgh&=\frac{1}{2}mv^2+mg\cdot 0 \\mh&=\frac{1}{2}v^2 \\v^2 &= 2gh \end{aligned}$

Conservation of energy when and after the ball hits the ground

$\begin{aligned} \frac{1}{2}mv'^2+mg\cdot 0&=\frac{1}{2}m\cdot 0+mgh' \\\frac{1}{2}mv'^2&=mgh' \\v'^2 &= 2gh' \\(0.8v)^2&=0.64v^2 = 0.64\times 2gh = 2gh' \\h'&=0.64h \end{aligned}$

Answer: D

Quantum Mechanics - Infinite Potential Well

See Problem #51

A measurement of energy E will always satisfy which of the following relationships?

A. E ≤ π²ħ²/8ma²

B. E ≥ π²ħ²/2ma²
C. E = π²ħ²/8ma²
D. E = n²π²ħ²/8ma²
E. E = π²ħ²/2ma²

(GR9277 #53)

Solution:

Wave function, $\psi_n=\sqrt{\frac{2}{a}} \sin \frac{n\pi x}{a}$
Time Independent Schrodinger Equation: $-\frac{\hbar^{2}}{2m}\psi''+ V\psi=E\psi$

Inside the well, V(x) = 0, and

$\\\psi=A\sin kx \\\psi'=Ak\cos kx \\\psi''=-Ak^2\sin kx$

$\\-\frac{\hbar^{2}}{2m}(-Ak^2 \sin kx)=EA\sin kx\rightarrow E=\frac{\hbar^{2}k^2}{2m}$

$\\k=\frac{n\pi}{a}\rightarrow E=\frac{n^2\pi^2 \hbar^{2}}{2ma^2} \\\\n=1,2,3, ...\rightarrow E\geqslant \frac{\pi^2 \hbar^{2}}{2ma^2}$

Answer: B

Thermal Physics - Heat Capacity

Questions 71-73

A system in thermal equilibrium at temperature T consists of a large number N₀ of subsystems, each of which can exist only in two states of energy E₁ and E₂, where $E_2-E_1=\epsilon>0$ . In the expressions that follow, k is the Boltzmann constant.

The internal energy of this system at any temperature T is given by $E_1 N_0+\frac{N_0\epsilon}{1+e^{\epsilon/kT}}$ . The heat capacity of the system is given by which of the following expressions?

A. $N_0 k \left (\frac{\epsilon}{kT} \right )^2\frac{e^{\epsilon/kT}}{\left (1+e^{\epsilon/kT} \right )^2}$

B. $N_0 k \left (\frac{\epsilon}{kT} \right )^2\frac{1}{\left (1+e^{\epsilon/kT} \right )^2}$

C. $N_0 k\left( \frac{\epsilon}{kT} \right )^2e^{-\epsilon/kT}$

D. $\frac{N_0 k}{2} \left (\frac{\epsilon}{kT} \right )^2$

E. $\frac{3}{2} N_0 k$

(GR9277 #72)

Solution:

Heat Capacity (at constant volume), $C_V=\left (\frac{\partial U}{\partial T} \right )_V$

$\\U= E_1 N_0+\frac{N_0\epsilon}{1+e^{\epsilon/kT}} \\\\\frac{\partial U}{\partial T}= N_0 k \left (\frac{\epsilon}{kT} \right )^2\frac{e^{\epsilon/kT}}{\left (1+e^{\epsilon/kT} \right )^2}$

Answer: A

Classical Mechanics - Circular Motion

A particle of mass M moves in a circular orbit of radius r around a fixed point under the influence of an attractive force F = ^K⁄_r_³, where K is a constant. If the potential energy of the particle is zero at an infinite distance from the force center, the total energy of the particle in the circular orbit is

A. −^K⁄_r²
B. −^K⁄_2r²
C. 0
D. ^K⁄_2r²
E. ^K⁄_r²

(GR9277 #87)

Solution:

Attractive force = Centripetal Force
^K⁄_r_³ = ^mv²⁄_r
mv² = ^K⁄_r²

Kinetic energy: T = ½mv² = ^K⁄_2r²
Potential energy: V(r) = − ∫ F dr = −K ∫ ¹⁄_r_³ dr = ^K⁄_2r²
Attractive force → negative potential energy, V(r) = − ^K⁄_2r²
Total energy: T + V = ^K⁄_2r² − ^K⁄_2r² = 0

Answer: C

Quantum Mechanics - Harmonic Oscillator

The energy levels for the one-dimensional harmonic oscillator are hν(n + ½), n = 0,1,2,⋯ How will the energy levels for the potential shown in the graph above differ from those for the harmonic oscillator?

A. The term ¹⁄₂ will be changed to ³⁄₂
B. The energy of each level will be doubled.
C. The energy of each level will be halved.
D. Only those for even values of n will be present.
E. Only those for odd values of n will be present.

(GR9277 #89)

Solution:

See GR9677 #98

Answer: E

Nuclear & Particle Physics - Hydrogen

The spacing of the rotational energy levels for the hydrogen molecule H₂ is most nearly

A. 10⁻⁹ eV
B. 10⁻³ eV
C. 10 eV
D. 10 MeV
E. 100 MeV

(GR9277 #90)

Solution:

Rotational kinetic energy: E = ^L²∕_2I
Angular momentum: L² = l(l+1)ħ² with l = 0,1,2,⋯
Moment Inertia: I = mr²

For hydrogen atom:
r = 0.529 × 10⁻¹⁰ m
m =10⁻²⁷ kg (mass of proton)
→ I = 10⁻²⁷(0.529×10⁻¹⁰)² ≈ 10⁻⁴⁷ kgm²
ħ = ^h∕_2π = ^{6.63×10⁻³⁴}∕ _2(3.14) ≈ 10⁻³⁴

For l = 0 → L = 0 → E = 0
For l = 1 → L = 2ħ² → E = ^ħ²∕_I

The spacing of the rotational energy levels:
∆E = ^ħ²∕_I − 0 = ^{(10⁻³⁴)²}∕ _10⁻⁴⁷ = 10⁻⁶⁸⁺⁴⁷ = 10⁻²¹ J

Convert to eV: 1 eV = 1.6 × 10⁻¹⁹ J
→ ∆E = ^10⁻²¹∕ _{1.6 × 10⁻¹⁹} ≈ 10⁻³eV

Answer: B

Thermal Physics - Specific Heat

For an ideal gas, the specific heat at constant pressure C_p is greater than the specific heat at constant volume C_v because the

Gas does work on its environment when its pressure remains constant while its temperature is increased.
Heat input per degree increase in temperature is the same in processes for which either the pressure or the volume is kept constant.
Pressure of the gas remains constant when its temperature remains constant.
Increase in the gas’s internal energy is greater when the pressure remains constant than when the volume remains constant
Heat needed is greater when the volume remains constant than when the pressure remains constant.

(GR8677 #14)

Solution:

(A) TRUE.
Heat Capacity: C = Q/dT

First law of Thermodynamics: the change in internal energy of a system dU is equal to the heat Q added and the work, W done on or by the system
→ dU = Q ± W

W done on the system → +W
W done by the system → −W

Gas (the system) does work on its environment
W done by the system
dU = Q − W

At constant V:
Work, W = PdV = 0
Q = dU
C_v = dU/dT

At constant P:
Work, W = PdV ≠ 0
Q = dU + W
C_p = dU/dT + PdV/dT = C_v + PdV/dT
C_p

C_v

(B) FALSE.
This means C_p = C_v, but according to A, C_p

C_v

(C) FALSE.
Ideal gas law: PV = NkT
If T constant, P changes if V changes.

(D) FALSE.
Heat Capacity, C = Q/dT does not depend on the gas’ internal energy, U

(E) FALSE.
See A. At constant V, Q = dU.
At constant P, Q = dU + W.

Answer: A

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 - Hydrogen

The energy levels of the hydrogen atom are given in terms of the principal quantum number n and a positive constant A by the expression

A. A(n + ½)
B. A(1 − n²)
C. A(¼ + 1/n)
D. An²
E. −A/n²

(GR8677 #19 )

Solution:

Energy level of hydrogen atom: E_n = −13.6/n² eV

Answer: E

Nuclear & Particle Physics - Photoelectric

Questions 31-33 refer to the apparatus used to study the photoelectric effect (see GR8677 #31).

The photoelectric equation is derived under the assumption that

Electrons are restricted to orbits of angular momentum nħ, where n is an integer
Electrons are associated with waves of wavelength λ = h/p, where p is momentum
Light is emitted only when electrons jump between orbits
Light is absorbed in quanta of energy E = hv
Light behaves like a wave

(GR8677 #32)

Solution:

According to the classical Maxwell wave theory of light, the average energy carried by an emitted electron should increase with the intensity of the incident light.

However, in photoelectric case, the energies of the emitted electrons are independent of the intensity of the incident radiation.

Einstein resolved this paradox by proposed that the incident light consisted of individual quanta, called photons, that interacted with the electrons in the metal like discrete particles, rather than as continuous waves.

Answer: D

Classical Mechanics - Conservative Force

A. ½mv²
B. −4kx³
C. kx⁴
D. −kx⁵/5
E. mg

(GR8677 #34)

Solution:

$F=-\nabla U =-\frac{\partial U}{\partial x} = -\frac{\partial}{\partial x}kx^4=-4kx^3$

Answer: B

Classical Mechanics - Hamiltonian

A. (p²/2m) + kx⁴
B. (p²/2m) − kx⁴
C. kx⁴
D. ½mv² − kx⁴
E. ½mv²

(GR8677 #35)

Solution:

Hamiltonian: H = T + U
U = kx⁴
T = ½mv² = p²/2m
H = (p²/2m) + kx⁴

Answer: A

Classical Mechanics - Lagrangian and Hamiltonian

Question 34-36

The potential energy of a body constrained to move on a straight line is kx⁴ where k is a constant. The position of the body is x, its speed v, its linear momentum p, and its mass m.

The body moves from x₁ at time t₁ to x₂ at time t₂. Which of the following quantities is an extremum for the x-t curve corresponding to this motion, if end points are fixed?

A. $\int_{t_1}^{t_2}\bigg(\frac{1}{2}mv^2-kx^4\bigg)dt$
B. $\int_{t_1}^{t_2}\frac{1}{2}mv^2dt$
C. $\int_{t_1}^{t_2}mxv$ $dt$
D. $\int_{x_1}^{x_2}\bigg(\frac{1}{2}mv^2+kx^4\bigg)dx$
E. $\int_{x_1}^{x_2}mv$ $dx$

(GR8677 #36)

Solution:

$\\$Lagrangian: $L=T-U \\$Hamiltonian: $ H=\int_{t_1}^{t_2} L dt \\U=kx^4 \\\\T=\frac{1}{2} mv^2 \\\rightarrow H=\int_{t_1}^{t_2}\bigg(\frac{1}{2} mv^2-kx^4\bigg)dt$

Answer: A

Thermal Physics - Blackbody Radiation

A blackbody at temperature T₁ radiates energy at a power level of 10 milliwatts (mW). The same blackbody, when at a temperature 2T₁, radiates energy at a power level of

A. 10 mW
B. 20 mW
C. 40 mW
D. 80 mW
E. 160 mW

(GR8677 #46)

Solution:

Blackbody radiation (Stefan-Boltzmann's law): u = σT⁴

Constant σ = u / T⁴

u₁T₁⁴ = u₂T₂⁴
10 / T₁⁴= u₂/ (2T₁)⁴
u₂ = (10)(2T₁)⁴/ T₁⁴= (10)(16) =160

Answer: E

Thermal Physics - Einstein solid

One feature common to both the Debye theory and the Einstein theory of the specific heat of a crystal composed of N identical atoms is that the

Average energy of each atom is 3kT
Vibrational energy of the crystal is equivalent to the energy of 3N independent harmonic oscillators
Crystal is assumed to be continuous for all elastic waves
Speed of the longitudinal elastic waves is less than the speed of the transverse elastic waves
Upper cutoff frequency of the elastic waves is the same

(GR8677 #51)

Solution:

Einstein solid:

Thermal properties of crystal
Highly idealized model focusing only the vibrational modes of crystal
Assumption: Each of N atoms is free to vibrate around its equilibrium position in any of 3 coordinate directions by a harmonic force with a natural frequency ω₀.

Einstein solid equipartition energy:

Number of oscillator N = N₀,
3-D Harmonic Oscillator, f = 6
U = (6/2) N₀kT = 3N₀kT

Answer: B

Thermal Physics - Conduction Electrons

The mean kinetic energy of electrons in metals at room temperature is usually many times the thermal energy kT. Which of the following can best be used to explain this fact?

A. The energy-time uncertainty relation
B. The Pauli exclusion principle
C. The degeneracy of the energy levels
D. The Born approximation
E. The wave-particle duality

(GR8677 #55)

Solution:

The mean kinetic energy of electrons in metals at room temperature is usually many times the thermal energy kT due to Pauli’s exclusion principle.

See problem GR0177 #76

Answer: B

Thermal Physics - Temperature

For a system in which the number of particles is fixed, the reciprocal of the Kelvin temperature T is given by which of the following derivatives? (Let P = pressure, V = volume, S = entropy, and U = internal energy.)

A. (∂P/∂V)_S
B. (∂P/∂S)_V
C. (∂S/∂P)_U
D. (∂V/∂P)_U
E. (∂S/∂U)_V

(GR8677 #66)

Solution:

The statistical mechanics definition of the temperature:

$T=\left( \frac{\partial S}{\partial U}\right )_{V,N}^{-1}$

The reciprocal of T:

$\frac{1}{T}=\left( \frac{\partial S}{\partial U}\right )_{V,N}$

Answer: E

Thermal Physics - Maxwell-Boltzmann Statistics

A large isolated of N weakly interacting particles is in thermal equilibrium. Each particle has only 3 possible non degenerate states of energies 0, ε, and 3ε. When the system is at an absolute temperature T ≫ ε/k, where k is Boltzmann’s constant, the average energy of each particle is

A. 0
B. ε
C. 4ε /3
D. 2ε
E. 3ε

(GR8677 #67)

Solution:

In thermal equilibrium:
→ Each particle has roughly equal probability of being in any of the three states.
→ The average energy of each particle is total energy/3 = (0+ε+3ε)/3 = 4ε/3

Answer: C

Note:

Complete Calculation:

T ≫ ε/k → 1 ≫ ε/kT
ε/kT → 0
e^-ε/kT → e⁰ = 1

Partition function: Z = ∑_i e^-ε_i/kT
i = 3 → Z = 3

Probability: P_i = e^-ε_i/kT/Z
→ P₁ = P₂ = P₃ = 1/3

Energy: E = ∑_iE_iP_i
→ E = E₁P₁+ E₂P₂+ E₃P₃ = (0+ε+3ε)/3 = 4ε/3

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