Cuymaca College Physics Internuclear Force Worksheet
ANSWER
Problem 1: Bond energy and zero-point vibrational energy
The lowest energy that a molecule can have due to its vibrational motion is known as the zero-point vibrational energy (ZPVE), which corresponds to the ground vibrational level (n = 0). The following is the formula for zero-point vibrational energy:
EZPVE=12ℏωEZPVE=21ℏω
Where:
The zero-point vibrational energy is abbreviated as EZPVE.
The decreased Planck constant has the value of (1.05457181034 Js1.05457181034Js).
The vibration’s angular frequency is given by the equation =km=mk.
, where mm is the decreased mass of the molecule, and kk is the force constant.
H2H2’s bond energy is listed as 4.48 eV and 4.48 eV.
Let us figure out the ZPVE:
Given:
K = 576 N/m K= 576 N/m
The mass of an H2H2 molecule is 21.6701027 kg.
To calculate:
=km = 576 N/m2 + 1.67 1027 kg =mk =2 =576 N/m
ω≈4.16×1013 rad/sω≈4.16×1013rad/s
Determine ZPVE:
EZPVE=12=12.0045718.01034 Js and 4.16 1013 rad/sEZPVE=21ℏω=21×1.0545718×10−34J⋅s×4.16×1013rad/s
EZPVE≈2.08×10−20JEZPVE≈2.08×10−20J
Let us now evaluate this energy about bond energy:
4.48 eV is reported as the bond energy. Convert eVeV to JJ in order to energies: 1 eV is equal equals 10-19 J.
This means that 4.48 eV = 4.48 1.60218 1019 J.
As 4.481.602181019>2.081020 J, the bond energy is larger than the ZPVE in magnitude.4.48×1.60218×10−19>2.08×10−20J.
Second issue: 57Co Decay
The average lifespan and decay constant
The half-life T1/2T1/2 and the decay constant are related by the formula =ln(2)T1/2=T1/2 ln(2).
Given:
T1/2=272 daysT1/2=272days
To calculate:
λ=ln(2)T1/2=ln(2)272 24 60 60 s = T1/2 ln(2) =272 24 60 60 sln(2)
Determine the average lifespan:
τ=1λτ=λ1
(b) The number of atoms
A radioactive source’s activity AA is calculated using the formula A=NA=N, where NN is the number of nuclei.
Given:
Activity A=2.00 mCi=2.00 106 Ci=2.00 1063.71010 s1A=2.00 mCi=2.00 106 Ci=2.00 1063.71010 s1 (conversion: 1 Ci=3.71010 s11Ci=3.71010s1) (calculated above).
Determine NN:
N=AλN=λA
(c) Activity One Year Later
A radioactive source’s activity AtAt is determined by the formula At=A0etAt =A0 et, where A0A0 represents the beginning activity.
Given:
T=1 year = 365 days, 60 minutes, and 60 seconds.
Initial endeavor A0 equals 2.00 106 3.7 1010 s1 (as determined above).
Determine AtAt:
At=A0e−λtAt=A0e−λt
Let us run the computations using the figures determined in the earlier step
QUESTION
Description
1. The force constant for the internuclear force in a hydrogen molecule 1H22 is k? = 576 N/m. A hydrogen atom has mass 1.67 x 10-27 kg. Calculate the zero-point vibrational energy for H2
(that is, the vibrational energy the molecule has in the n = 0 ground vibrational level). How does this energy compare in magnitude with the H2 bond energy of -4.48 eV?
2. The isotope 57Co decays by electron capture to 57Fe with a half- life of 272 days. The 57Fe nucleus is produced in an excited state, and it almost instantaneously emits gamma rays that we
can detect. (a) Find the mean lifetime and decay constant for 57Co. (b) If the activity of a 57Co
radiation source is now 2.00 mCi, how many 57Co nuclei does the source contain? (c) What will be the activity after one year?
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