1

Physics 2102B – 2nd Assignment

Your solution to problems 1, 7, 8, and 9 has to be handed in, in class, on Monday, Mars 3, 2014.

1. Consider several alpha particles approaching a nucleus, with various impact parameters b , but all with the same total energy E . Prove that an alpha particle that approaches the nucleus head-on (b = 0 ) gets closer to the nucleus than any other.

2. (Prob. 18, Ch. 4, in Thornton and Rex.) Calculate the time, according to classical laws, it would take the electron of a hydrogen atom to radiate its energy and crash into the nucleus. [Hint: The radiated power P is given by

P = 14πε0

2e2

3c3d 2rdt 2

⎛⎝⎜

⎞⎠⎟

2

, (2.1)

where Q is the charge, c the speed of light, and r the position vector of the electron from the centre of the atom. Apply the conservation of energy.] Solution. The total energy of the atom is (see equation (3.43) in Chap. 3 of the lecture notes)

E = − 12

e2

4πε0r, (2.2)

and its time derivative

dEdt

= 12

e2

4πε0r2drdt. (2.3)

But by conservation of energy this derivative is equal to minus the radiated power, and using equation (2.1)

12

e2

4πε0r2drdt

= − e2

4πε023c3

d 2rdt 2

⎛⎝⎜

⎞⎠⎟

2

, (2.4)

or

2

drdt

= − 4r 2

3c3

d 2rdt2

⎛⎝⎜

⎞⎠⎟

2

. (2.5)

However for a (classical) hydrogen atom, the acceleration d

2r dt2 of the electron on its (circular) orbit is simply the centripetal acceleration due to the electrostatic force

d 2rdt2 = − e2

4πε0mr 2 (2.6)

Inserting this relation into equation (2.5) we find

drdt

= − e2

4πε0

⎛⎝⎜

⎞⎠⎟

24

3m2c3r2, (2.7)

and

t = − 3m2c3

4e2

4πε0

⎛⎝⎜

⎞⎠⎟

−2

r2 dra0

0

∫

= − m2c3

4e2

4πε0

⎛⎝⎜

⎞⎠⎟

−2

r3a0

0

= 1.55 ×10−11s.

(2.8)

3. (Prob. 53, Ch. 4, in Thornton and Rex.) The proton (of mass M ) and the electron (of mass m ) in a hydrogen atom actually rotate about their common centre of mass. The distance r = re + rM is still defined as the electron-nucleus distance (see the figure on the right). Show that the equation for the radius of Bohr’s stationary states

rn =

4πε0n22

µe2, (3.1)

for n = 0, 1, 2,… and e the elementary charge, is only modified by substituting m by the reduced mass

µ = MmM +m

. (3.2)

3

[Hint: You can simplify your calculations by setting the centre of mass of the hydrogen atom to zero.] Solution. If we set the centre of mass R = rem − rMM( ) M +m( ) to zero ( re and rM are greater than zero, by definition) we get

rM = mM

re

r = re + rM

= re 1+ mM

⎛⎝⎜

⎞⎠⎟

,

(3.3)

and

re =M

M + mr

rM = mM + m

r. (3.4)

But by Newton’s third law the centrifugal force on the nucleus and electron must be equal in magnitude (since they counterbalance the electrostatic force, which is the same for both the electron and the nucleus) and

mve2

re

=MvM

2

rM

, (3.5)

or, from equations (3.4),

vM

ve

= mM

. (3.6)

Now the total angular momentum of the system is

L = mvere + MvM rM

= mvere + M mM

ve

⎛⎝⎜

⎞⎠⎟

mM

re

⎛⎝⎜

⎞⎠⎟

= mver,

(3.7)

and since from quantization we also have L = n then

4

ve =

nmr. (3.8)

We finally equate the centrifugal and electrostatic forces on the electron

e2

4πε0r2 =

mve2

re

= m nmr

⎛⎝⎜

⎞⎠⎟

2M + m

Mr⎛⎝⎜

⎞⎠⎟

= n22

µr3 ,

(3.9)

or

rn =

4πε0n22

µe2. (3.10)

4. We know that postulating the existence of a de Broglie wave for the orbiting electron of a hydrogen atom, with the assumption that stationary states are achieved when conditions leading to standing waves are verified, naturally leads to the quantization of the orbital angular momentum (as previously postulated by Bohr). Using this quantization as a starting point, show that the kinetic energy of the hydrogen atom is given by

K = 12nhforb , (4.1)

with n = 1, 2, 3,… and forb the orbital frequency of the electron about the nucleus. Solution. For a classical (circular) orbit we have for the angular momentum

L = mvr

= mv v2π forb

⎛⎝⎜

⎞⎠⎟

= Kπ forb

,

(4.2)

where K is the kinetic energy. But we also assume that L = n , which implies that

5

K = π forbn =

12nhforb . (4.3)

5. Let us start with the general relation vph = aλn , with a and n some constants, for the

phase velocity of waves in shallow water. It is found for this kind of waves that the phase velocity is equal to the group velocity. What is the dependence of the phase velocity on the wavelength? Solution. We start with ω = kvph and calculate

vg =dωdk

= vph + kdvphdk

= vph + kdvphdλ

dλdk

= aλ n + k anλ n−1( ) − 2πk2

⎛⎝⎜

⎞⎠⎟

= aλ n 1− n( ).

(5.1)

But since vph = vg we find that n = 0 , and the phase velocity is independent of the wavelength.

6. (Prob. 54, Ch. 5, in Thornton and Rex.) Consider a wave packet having the product Δp ⋅ Δx = at a time t = 0 . What will be the spatial width of such a wave packet after the

time m Δx( )2 ? Solution. At time t = 0 the uncertainty in velocity is

Δv0 =Δpm

= mΔx

. (6.1)

After a time m Δx( )2 the wave packet will have travelled a distance

6

′x ≈ v0

m Δx( )2

⎡

⎣⎢⎢

⎤

⎦⎥⎥, (6.2)

which implies that it will have accrued a new uncertainty

Δ ′x ≈ Δv0m Δx( )2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

≈ Δx.

(6.3)

Since uncertainties add quadratically the final spatial width of the wave packet will be

Δxf ≈ Δx( )2 + Δ ′x( )2

≈ 2Δx. (6.4)

7. Model the electron of a hydrogen atom as a simple (quantum mechanical) harmonic oscillator. That is, assume that the electron of mass m is bound to the nucleus through a spring of constant κ , as it evolves on a circular orbit. Consider the total energy of the oscillator and use the Heisenberg uncertainty to show that its minimum energy is ω 2 , where ω = 2πν with ν the frequency of oscillation.

8. (Prob. 65, Ch. 5, in Thornton and Rex.) Aliens visiting Earth are fascinated by baseball. They are so advanced that they have learned how to vary ! to make sure that the pitcher cannot throw a strike with any confidence. Assume that the width of the strike zone is Δx = 0.38 m , the speed of the baseball is v = 35 m/s , the mass of the baseball is 145 g , and the ball travels a distance of d = 18 m from the pitcher’s hand to the strike zone. What is the minimum value of ! required to prevent the pitcher from confidently throwing a strike? [Hints: There are two uncertainties involved in this problem: for the width of the strike zone and the transverse momentum of the baseball.]

9. Consider a particle confined in a one-dimensional box of length L . Use the Heisenberg uncertainty to show that the particle can never have zero kinetic energy. Determine the minimum level of kinetic energy for the non-relativistic case.

10. Consider a Gaussian wave function at time t = 0 of the form

7

ψ x,0( ) = Ae− x2Δx

⎛⎝⎜

⎞⎠⎟2

e jk0x , (10.1) where A is a constant and the spatial extent of the envelope is set by Δx . Calculate the Fourier transform ψ k( ) of equation (10.1) and show that it also has a Gaussian form with a spectral with Δk = 1 2Δx( ) . Is this result consistent with the Heisenberg inequality? Solution. Let us calculate the Fourier transform

ψ k( ) = 12π

ψ x( )e− jkx dx−∞

∞

∫

= 12π

Ae− x2Δx

⎛⎝⎜

⎞⎠⎟2

e jk0xe− jkx dx−∞

∞

∫

= A2π

e− x2Δx

⎛⎝⎜

⎞⎠⎟2

e− j k−k0( )x dx−∞

∞

∫

= A2π

e− Δx⋅ ′k( )2 e− 12Δx( )2

x+ j2 Δx( )2 ′k⎡⎣

⎤⎦2

dx−∞

∞

∫

(10.2)

where we have defined ′k = k − k0 . We now make the following change of variable

′x = x + j2 Δx( )2 Δk and equation (10.2) becomes

ψ k( ) = A2π

e− Δx⋅ ′k( )2 e− ′x2Δx

⎛⎝⎜

⎞⎠⎟2

d ′x−∞

∞

∫

= A2π

e− Δx⋅ ′k( )2 ⋅2Δx π

= A 2Δxe− Δx⋅ ′k( )2 ,

(10.3)

where we solved the integral using the result of Problem 1 in the first assignment. If we define

ψ k( ) = Be− ′k2Δk

⎛⎝⎜

⎞⎠⎟2

, (10.4) then we find Δk = 1 2Δx through a comparison with the last of equations (10.3), which implies that

8

Δk ⋅ Δx = 12. (10.5)

This is the minimum value allowed by the Heisenberg inequality, and is therefore consistent with it.