Physics Problems & Solutions: #97

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Quantum Mechanics - Probability Current Density

A particle of mass m has the wave function

ψ(x,t) = e^iωt [αcos(kx) + βsin(kx)]

where α and β are complex constant and ω and k are real constants. The probability current density is equal to which of the following? (Note: α* denotes complex conjugates of α, and |α|²= α*α .)

(GR9677 #97)

Solution:

Probability density function:

$J = \frac{\hbar}{2mi} \left( \psi^* \nabla \psi - \psi \nabla \psi^*\right)<br><br>$

Answer: E

Condensed Matter - Effective mass

Lattice forces affect the motion of electrons in a metallic crystal, so that the relationship between the energy E and wave number k is not the classical equation E = ħ²k²/2m, where m is the electron mass. Instead, it is possible to use an effective mass m* given by which of the following?

A. $m^*=\frac{1}{2}\hbar^2 k \left( \frac{dk}{dE} \right)$

B. $m^*= \frac{\hbar^2 k^2}{\left( \frac{dk}{dE} \right)}$

C. $m^*= \frac{1}{2} \hbar^2 m^2 \left( \frac{d^2k}{dE^2} \right)^{1/3}$

D. $m^*= \frac{\hbar^2}{\left( \frac{d^2E}{dk^2} \right)}$

E. $m^*=\frac{1}{2} \hbar^2 m^2 \left( \frac{d^2E}{dk^2} \right)$

(GR9277 #97)

Solution:

F = m*a

F = dp/dt = ħ dk/dt (momentum, p = ħk)

a = dv_g/dt

Group velocity, v_g= dω/dk

Energy of electron, E = hf = ħω
ω = E/ħ

$v_g=\frac{d\omega}{dk} = \frac{1}{\hbar} \left( \frac{dE}{dk} \right)$

$a=\frac{dv_g}{dt}= \frac{1}{\hbar} \frac{d}{dt}\left( \frac{dE}{dk} \right)\frac{dk}{dk}= \frac{1}{\hbar} \left( \frac{d^2E}{dk^2} \right)\frac{dk}{dt}$

F = m*a

$\hbar \frac{dk}{dt} =m^*\cdot \frac{1}{\hbar} \left( \frac{d^2E}{dk^2} \right)\frac{dk}{dt}$

$m^*= \frac{\hbar^2}{\left( \frac{d^2E}{dk^2} \right)}$

Classical Mechanics - Rotational Motion

Two uniform cylindrical disks of identical mass M, radius R, and moment inertia ½MR² collide on a frictionless, horizontal surface. Disk I, having an initial counterclockwise angular velocity ω₀ and a center-of-mass velocity v₀ =½ω₀R to the right, makes a grazing collision with disk II initially at rest. If after the collision the two disks stick together, the magnitude of the total angular momentum about the point P is

A. Zero
B. ½MR²ω₀
C. ½MR²v₀
D. MRv₀
E. Dependent on the time of the collision

(GR8677 #97)

Solution:

L_total = L_trans + L_rot

L_trans = r × p
r = distance from the center of disk I to point P.
At point P, R = r.

L_trans = R × p = R × Mv₀ = M(R × v₀)

From the problem, ω₀ is counterclockwise and v₀is to the right. Thus, the crossproduct (R × v₀) is negative. Also, v₀ =½ω₀R

L_trans = −½MR²ω₀

L_rot= Iω₀

Moment inertia, I = ½MR²

L_rot= ½MR²ω₀

L_total = −½MR²ω_{0 +}½MR²ω₀= 0

Answer: A

Optics - Snell's Law

A beam of light has a small wavelength spread δλ about a central wavelength λ. The beam travels in vacuum until it enters a glass plate at an angle θ relative to the normal to the plate. The index of refraction of the glass is given by n(λ). The angular spread δθ' is given by

A. $\delta \theta'=\left | \frac{1}{n}\delta\lambda \right |$

B. $\delta \theta'=\left | \frac{dn(\lambda)}{n}\delta\lambda \right |$

C. $\delta \theta'=\left | \frac{1}{\lambda}\frac{d\lambda}{dn}\delta\lambda \right |$

D. $\delta \theta'=\left | \frac{\sin \theta}{\sin \theta'}\frac{\delta \lambda}{\lambda} \right |$

E. $\delta \theta'=\left | \frac{\tan \theta'}{n}\frac{dn(\lambda)}{d\lambda} \delta \lambda\right |$

(GR0177 #97)

Solution:

Snell's Law: n₁sin θ₁ = n₂sin θ₂

Given:
n₁= 1 for vacuum
n₂= n(λ)
θ₁= θ
θ₂= θ'

sin θ = n(λ)sin θ'

Take the derivative of the equation with respect to λ:

^dsin θ/_dλ= ^dn^(λ)^sin θ'/_dλ ...(Eq.1)

θ is a constant → ^dsin θ/_dλ= 0

Eq.1 →
0 = ^dn^(λ)^sin θ'/_dλ
0 = n(λ) (^dsin θ'/_dλ) + sin θ' (^dn^(λ)/_dλ) ...(Eq.2)

Since n is a function of λ, θ' is also a function λ →
^dsin θ'/_dλ= (^dsin θ'/_dθ')(^dθ'/_dλ) = cos θ' (^dθ'/_dλ)

Eq.2 →
0 = n(λ) cos θ' (^dθ'/_dλ) + sin θ' (^dn^(λ)/_dλ)
n(λ) cos θ' (^dθ'/_dλ) = − sin θ' (^dn^(λ)/_dλ)
n(λ) (^dθ'/_dλ) = − tan θ' (^dn^(λ)/_dλ)
^dθ'/_dλ = − (^tan θ'/_n(_λ)) (^dn^(λ)/_dλ)
δθ' = |(^tan θ'/_n(_λ)) (^dn^(λ)/_dλ)|

Answer: E

Nuclear & Particle Physics - Compton Scattering

In the Compton effect, a photon with energy E scatters through 90^o angle from stationary electron mass m. The energy of the scattered photon is

A. E
B. E/2
C. E²/mc²
D. E²/(E + mc²)
E. E∙mc²/(E + mc²)

(GR0877 #97)

Solution:

Compton scattering: λ_f − λ_i = ^h/_mc(1− cosθ)
Photon energy, E = ^hc/_λ

^hc/_{E_f} − ^hc/_E = ^h/_mc(1 − cos 90^o)
¹/_{E_f} − ¹/_E = ¹/_mc_²(1 − 0)
¹/_{E_f}  = ¹/_m_c_²+ ¹/_E
^Em^c²/_{E_f}  = E + mc²
^Em^c²/_{E_f}  = E + mc²
E_f = ^Em^c²/_{E +}_mc_²

Answer: E

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