Chem1000A
Fall 2003
Assignment 10 - Solutions
Two or three questions will be marked from each assignment.
DUE ON Nov. 21, 2003 (Friday) 1:00 PM
To be dropped off at my office (C886)
1. One line in the atomic emission spectrum of hydrogen has a wavelength of 656.3 nm.
(a) Calculate the frequency of this radiation.
ν = c/ λ = 2.9979 × 108 m s-1/656.3 × 109 m = 4.568 × 1014 s-1= 4.568 × 1014 Hz
(b) Calculate the energy of one photon.
E = h ν = 6.626 × 10-34 Js × 4.568 × 1014 s-1 = 3.027 ×10-19 J
(c) Calculate the energy of one mole of photons.
Molar energy = Emolar = 3.027 ×10-19 J× 6.022 × 1023 mol-1=182300 J/mol = 182.3 kJ/mol
(d) What colour does this emission have?
Red light is emitted.
2. Calculate the energies (∆E has a sign), the wavelengths (λ has no sign), and the frequencies (ν has
no sign) of the following transitions. Specify whether the transition involves emission or absorption
of light; argue using the sign of ∆E.
(a) The transition of one electron in He+ from n = 2 to n = 4.
For He: Z=2
∆E = Efinal – Einitial = -Ry (4/42 – 4/22)
= - 2.179 × 10-18 J ×(4/16 – 4/4) = - 2.179 × 10-18 J ×( – 3/4)
= 1.634 × 10-18 J
A photon of the energy 1.634 × 10-18 J is absorbed, since ∆E is positive (endothermic process).
ν = E/h = 1.634 × 10-18 J/(6.626 × 10-34 Js) = 2.466 × 1015 s-1 = 2.466 × 1015 Hz
λ = c/ν = 2.9979 × 108 m s-1/(2.466 × 1015 s-1) = 1.215 × 10-7 m = 1.215 × 102× 10-9 m = 121.5 nm
(b) The transition of one electron in H from the K shell to the M shell.
For H: Z=1; K-shell: n = 1; M shell: n = 3
∆E = Efinal – Einitial = -Ry (1/32 – 1/12)
= - 2.179 × 10-18 J ×(1/9 – 1) = - 2.179 × 10-18 J ×( – 8/9)
= 1.937 × 10-18 J
A photon of the energy 1.937 × 10-18 J is absorbed, since ∆E is positive (endothermic process).
ν = E/h = 1.937 × 10-18 J/(6.626 × 10-34 Js) = 2.923 × 1015 s-1 = 2.923 × 1015 Hz
λ = c/ν = 2.9979 × 108 m s-1/(2.923 × 1015 s-1) = 1.026 × 10-7 m = 1.026 × 102× 10-9 m = 102.6 nm
(c) The ionization of He+ (in the ground state) producing He2+.
For He: Z=2
∆E = Efinal – Einitial = -Ry (4/∞2 – 4/12) = - 2.179 × 10-18 J ×(0 – 4) = - 2.179 × 10-18 J ×( – 4)
= 8.716 × 10-18 J
A photon of the energy 8.716 × 10-18 J is absorbed, since ∆E is positive (endothermic process).
ν = E/h = 8.716 × 10-18 J/(6.626 × 10-34 Js) = 1.315 × 1016 s-1 = 1.315 × 1016 Hz
λ = c/ν = 2.9979 × 108 m s-1/(1.315 × 1016 s-1) = 2.279 × 10-8 m = 2.279 × 101× 10-9 m = 22.79 nm
1
3. To cause a cesium atom on a metal surface to lose an electron, an energy of 2.0 × 102 kJ mol-1 is
required. (a) Calculate the longest possible wavelength that can ionize a cesium atom. (b)What type
of electromagnetic radiation is necessary to ionize Cs atoms? (c) If a wavelength of 400 nm is used,
what is the kinetic energy of the photoelectron?
(a) energy required for one atom instead of 1 mol atoms:
2.0 × 102 kJ mol-1/6.022 × 1023 mol-1 = 2.0 × 105 J mol-1/6.022 × 1023 mol-1 = 3.3 × 10-19 J
E = hν
ν = E/h = 3.3 × 10-19 J/(6.626 × 10-34 Js) = 5.0 × 1014 s-1 = 5.0 × 1014 Hz
c = λν
λ = c/ν = 2.998 × 108 m s-1/(5.0× 1014 s-1) = 6.0 × 10-7 m = 6.0 × 102× 10-9 m = 600 nm
(b) This radiation is found in the visible region of the spectrum (orange).
(c) c = λν
ν = c/λ = 2.998 × 108 m s-1/(400× 10-9 m) = 7.50 × 1014 s-1
E = hν = 6.626 × 10-34 Js × 7.50 × 1014 s-1 = 4.97 × 10-19 J
Ekin = E400nm - Eionization = 4.97 × 10-19 J - 3.3 × 10-19 J = 1.7 × 10-19 J
Ekin = 1.7 × 10-19 J × 6.022 × 1023 mol-1 = 100000 J mol-1 = 1.0 × 102 kJ mol-1
4. Calculate the wavelength associated with a neutron that has a kinetic energy of 6.21 × 10-21 J. The
mass of a neutron is 1.675 × 10-24 g.
Ekin = m/2 v2; v = (2Ekin/m)1/2
v = (2 ×6.21 × 10-21 J/(1.675 × 10-24 g))1/2 = (2 ×6.21 × 10-21 kg m2 s-2/(1.675 × 10-27 kg))1/2
= 2723 m/s
λ = h/(mv) = 6.626 × 10-34 Js/(1.675 × 10-27 kg ×2723 m/s) = 1.45× 10-10 (kg m2 s-2) s kg-1 m-1 s
= 1.45× 10-10 m = 0.145 nm
c
hc
m
m
; ρ = ; pV = nRT ; PT = ∑ pi ; pi = X i ⋅ PT ; E = hν =
; λ = ; E = mc 2 ;
λ
ν
V
M
i
 Z2
 Z2
h
1
Z2 
Z2 
E kin = mv 2 ; ∆E = (E final − E initial ) = − R Ryd hc 2 − 2  = − Ry 2 − 2  ; λ =
n

n

mv
2
 final ninitial 
 final ninitial 
Z2
h
∆x ⋅ ∆(mv ) >
; E n = − Ry 2
4π
n
-18
Ry = 2.179 × 10 J; h = 6.626 × 10-34 Js, c = 2.9979 × 108 m/s
n=
2
;