PH300
HW6 answer key
1) What is the cross-section (in barns) for alpha particle with kinetic
energy of 8 MeV to scatter by 90 degrees or more from a gold
nucleus? Consider a 2 mm diameter beam of 106 alpha particle per
second incident on a gold target. There are 5.0x1017 gold nuclei in
the beam area. How many alpha particles per second will scatter by
more than 90 degrees?
Using
Z1 Z2 e2
b=
cot θ
8π ε 0 K
2
where e /4πε = 1.44x10
2
0
-9
eV*m
and in this problem Z1 = 2, Z2 = 79, K = 8x106 eV, θ = 90˚
plugging in these values yields:
b = 1.422x10-14 m
The cross section is σ = πb2 = 6.35x-28 m2 = 6.35 b
The total scattering area is the area of the beam of alpha particles.
AT = πr2 = (3.14)*{(0.001 m)2} = 3.14x10-6 m2 or 3.14x1022 b
Effective area to scatter 90˚: (5x1017)*(6.35 b) = 3.176x1018 b
Probability to scatter 90˚ = (effective scattering area)/(total scattering area)
= 1.011x10-4
Intensity of scattered alpha particles:
(1.011x10-4)*106 ~ 101.1 alpha particles per second (90˚ scatter)
3) What is the speed (v/c) of an electron in the first three Bohr orbits
of the H atom? (Ch4 #22)
Vn = 1/n*/ma0 = 1/n*e2/4πε0 = 2.2x106/n or {1/(n*137)}*c
n =1: β = 1/(1*137) = 0.0073
n =2: β = 1/(2*137) = 0.0036
n =3: β = 1/(3*137) = 0.0024
3) A hydrogen atom in an excited state absorbs a photon of
wavelength 434 nm. What were the initial and final states of the
hydrogen atom? (ch 4# 23)
E = hc/λ = 1240 eV*nm/(434 nm) = 2.857 eV
Original level is 2.857 eV or greater, so that leaves n = 1 or 2
n = 1 → E =-13.6 eV
n = 2 → E =-3.40 eV
The initial level nint cannot be n =1 because an energy jump of 2.857 is not
allowed. nint must be 2.
Ef = Ei - ∆E; Ef = -3.40 eV +2.857 eV = -0.54 eV
Does 0.54 eV correspond to an acceptable energy level in the hydrogen atom?
Yes, level 5. So an electron in level 2 was excited when a photon was
absorbed by the hydrogen atom to level 5.
4) If a hydrogen atom is initially in the first excited state, what is the
longest wavelength of light it will absorb? What is the shortest
wavelength of light it will absorb?
The lowest excited state is n = 2. From problem #4 we know that this
corresponds to an energy of -3.40 eV. When a photon is absorbed and an
electron is excited to a higher energy state then the largest and smallest
wavelength photons correspond to the lowest and highest energy transitions
respectively.
Lowest energy photon that can be absorbed:
n = 2 jumps to n = 3
∆E = 3.40 eV – 1.51 eV = 1.89 eV
λlow = hc/∆E = 1240 eV*nm/ 1.89 eV = 656 nm
Highest energy photon that can be absorbed:
n = 2 jumps to n = ∞ (electron leaves nucleus and atom is ionized)
∆E = 3.40 eV – 0 eV = 3.40 eV
λhigh = hc/∆E = 1240 eV*nm/ 3.40 eV = 364.7 nm
5) What is the calculated binding energy of the electron in the ground
state of (a) deuterium, (b) He+, (c) Li++? (ch4# 25)
1
Starting with equation λ

=Z2 R
R=
reduced mass correction
1 1
−
n2l n2u
1
R
me ∞
1
M
where Z is the atomic number, using the
where me is the mass of an electron
(5.4858x10-4 amu), M is the mass of the nucleus and the Rydberg constant R∞
has a value of 0.010974 nm-1
and since this is a binding energy problem nl = 1 and nu = ∞
 
1
R∞
Also E = hc(1/λ) so E = (1240 eV*nm)*Z * 1me
M
2
deuterium: Z = 1, M = 2.01355 amu → E = 13.6037 eV
He+ :
Z = 2, M = 4.00151 amu → E = 54.4222 eV
Li++ :
Z = 3, M = 7.01436 amu → E = 122.457 eV
8)Extra credit 6-28
The gravitational force provides the centripetal acceleration v^2/r, só
F=GMeMp/r^2=Mev^2/r.
Manipulating this equation, we can find kinetic energy
K=Mev^2/2=GMeMp/(2r)--------(1).
The potential energy of the systerm is U=-GMeMp/r.
Thus E=K+U=-GMeMp/(2r)---------(2).
Angular momentum L=Mevr=nћ, só v=nћ/(Mer) and
Mev^2/2=Me(nћ/Mer)^2/2.
Then from equation (1), we can obtain Me(nћ/Mer)^2=GMeMp/r,
thus
rn=(nћ)^2/(GMpMe^2)=a0 n^2.
Só we got a0=ћ^2/(GMpMe^2)=1.2056*10^29m.
En=-GMeMp/(2rn)=-(GMeMp)^2 Me/(2 n^2 ћ^2).
Só the difference energy between n=2 and n=1 is
-(GMeMp)^2 Me/(2 ћ^2)(1/4-1)=1.968*10^-78ev=3.15*10^-97J