PHY6095/PHZ6166: Midterm: Solutions Wednesday, February 27

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PHY6095/PHZ6166: Midterm: Solutions
Wednesday, February 27
2
• Problem 1: 35 points
How strong a (static) electric field should be to ionize a hydrogen atom? Express your answer via the fundamental
constants: the electron charge e, the electron mass m, the Planck constant h̄, the speed of light c, the Boltzmann
constant kB , etc. (The list above is neither inclusive or exclusive).
An atom is ionized if work done by the electric field on the distance comparable to the bound state radius (Bohr
radius, aB = h̄2 /me2 in SGS) is comparable to the binding energy
eEaB ∼ e2 /ab → E ∼ e/a2B ∼
e5
.
m2 h̄4
• Problem 2: 30 points
Find the asymptotic behavior of
Z
J=
∞
dx
0
cos(ωx)
x3 + 1
for ω 1. The integrand is a non-analytic function of x (there is a pole on the real axis at x = −1), therefore,
we need to integrate by parts (IBP). A finite answer can only come from f (0) or some of its derivative f n (0).
One IBP gives sin(ωx)f (x) |∞
0 /ω = 0, therefore we need to integrate further. To see how many times we need to
do IBP, we look at the derivatives of f (x) = 1/(1 + x3 ) at x = 0. Since x will put equal to 0 after differentiation,
we can expand f (x) for small x and then differentiate:
f (x) ≈ 1 − x3 .
000
Therefore, the first non-vanishing derivative is f (0) = −6. Applying IBP four times, we get a boundary term
000
4
4
− cos(ωx)f (x) |∞
0 /ω = −6/ω .
• Problem 3: 35 points Function I(a, b) is defined as an integral
Z ∞
2
dx √
I(a, b) =
J0 ( x/b) ,
2
x+a
0
(1)
where a, b > 0 and J0 (x) is the Bessel function of order zero. Find the leading term in the asymptotic expansion
of I(a, b) for a b.
Hint: Recall that J0 (0) = 1.
Rescale y = x/b2 and integrate by parts
Z ∞
Z ∞
2 √ ∞
d 2 √
dy
2 √
2
J
(
y)
=
ln
y
+
(a/b)
J
(
y)
|
−
J ( y).
I=
ln y + (a/b)2
0
0
0
2
y + (a/b)
dy 0
0
0
The remaining integral is convergent for a/b = 0, so it becomes a number ∼ 1 in this limit. Therefore the
leading term is given by
b
I ≈ 2 ln .
a
• Bonus: 20 points
Find the leading term in the asymptotic expansion of the integral in Problem 3 for a b.
q
2
Hint: Recall that J0 (z) ≈ πz
sin(z + π4 ) for z 1.
Rescale y = x/a2 :
Z
I=
∞
dy
0
a√ 1
J02
y .
y+1
b
3
Since a/b 1, we replace J0 by its large-argument asymptotics:
Z
a√
1
1
π
2b ∞
dy
y+
I≈
√ sin2
πa 0
y+1 y
b
4
Z ∞
h
1b
1
a√
1
π i
=
dy
y+
√ 1 − cos 2
πa 0
y+1 y
b
2
The cos is rapidly oscillating and thus can be discarded. The remainder gives
Z
Z
1b ∞
1
b
1
2 b ∞ dz
I≈
dy
= .
√ =
πa 0
1+y y
π a 0 1 + z2
a
(2)
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