Answer:
The frequency of the sulphur is 2.6×10^13 Hz.
We find the frequency of the sulphur because sulphur is just below the oxygen atom in the periodic table. So, we find it by using the frequency of the oxygen.
Explanation:
As, the frequency in terms of spring constant and mass is
:
f = (1/2π) × √(k/m)
k is the spring constant and m is the mass.
Square on both sides of the equation and it becomes then,
f^2 = (1/4π^2) × (k/m)
f^2×m = k/4π^2
Now, we will solve it with oxygen and sulphur because the oxygen atom is chemically replaced with a sulphur atom, the spring constant of the bond is unchanged so, the relation between fo and fs is:
<h3>fo^2×mo = fs^2×ms</h3>
As, fo is the frequency of the oxygen and fs is the frequency of the sulphur atom and mo is the mass of the oxygen and ms is the mass of the sulphur.
So,
fs^2 = (fo^2×mo) / ms
fs = fo×√(mo/ms)
As, we also know the mass of the oxygen and sulphur which is 16amu and 32amu:
fs = (3.7×10^13) × √(16/32)
fs = (3.7×10^13) × 1/√2
fs = 2.6×10^13 Hz.
Therefore, the frequency is: 2.6×10^13 Hz.
