CM1111 Mass Tutorial 1
AY2017/2018 S1
1
Use the following constants for the questions below.
NA = 6.022 x 1023 mol-1; h = 6.6261 x 10-34 J s; c = 2.99792 x 108 ms-1; 1eV= 1.6022 x 10-19 J.
1. Use the values below to calculate the corresponding wavenumber of the photons in cm-1 (to 4
significant figures) and state where in the electromagnetic (EM) spectrum the respective
photons would lie. Refer to the EM spectrum plot on slide 13 of Lecture Note “CM
1111_1_Atomic structure_Part_1”.
a. 1 kJ mol-1
b. 1000 nm
c. 1 x 10-3 eV
Solution:
a. 1 kJ mol-1
We have 1000 J of energy per mole of photons. NA = 6.022 x 1023 mol-1, so we must have
(1000/ NA) J per photon = 1.661 x 10-21 J.
E = h = hc/, so 1/ = E/hc = 1.661 x 10-21/(6.6261 x 10-34 x 2.99792 x 108) = 8362 m-1=
8362 /(1m)= 8362 /(100cm)=83.62 cm-1.
Therefore 1 kJ mol-1 83.62 cm-1
1/ = 83.62 cm-1, so = (1/83.62) cm = 1.196 x 10-2 cm
– this is MW radiation.
b. 1000 nm
= 1000 nm = 1000 x 10-9 m = 1000 x 10-7 cm = 1.000 x 10-4 cm,
so 1/ = (1.000 x 104 ) cm-1
Therefore 1000 nm 1.000 x104 cm-1.
1000 nm = 1 x 10-4 cm – this is IR radiation.
