Answer:
a. the minimum amplitude of vibration for the NO molecule A [tex]\simeq[/tex] 4.9378 pm
b. the minimum amplitude of vibration for the HCl molecule A [tex]\simeq[/tex] 10.9336 pm
Explanation:
Given that:
The effective spring constant describing the potential energy of the HBr molecule is 410 N/m
The effective spring constant describing the potential energy of the NO molecule is 1530 N/m
To calculate the minimum amplitude of vibration for the NO molecule, we use the formula:
[tex]\dfrac{1}{2}kA^2= \dfrac{1}{2}hf[/tex]
[tex]kA^2= hf[/tex]
[tex]A^2= \dfrac{hf}{k}[/tex]
[tex]A = \sqrt{\dfrac{hf}{k}}[/tex]
[tex]A = \sqrt{\dfrac{(6.626 \times 10^{-34} \ J. s ) ( 5.63 \times 10^{13} \ s^{-1})}{1530 \ N/m}}[/tex]
[tex]A = \sqrt{\dfrac{3.730438 \times 10^{-20} \ m}{1530 }[/tex]
[tex]A = \sqrt{2.43819477 \times 10^{-23}\ m[/tex]
[tex]A =4.93780799 \times 10^{-12} \ m[/tex]
A [tex]\simeq[/tex] 4.9378 pm
The effective spring constant describing the potential energy of the HCl molecule is 480 N/m
To calculate the minimum amplitude using the same formula above, we have:
[tex]A = \sqrt{\dfrac{hf}{k}}[/tex]
[tex]A = \sqrt{\dfrac{(6.626 \times 10^{-34} \ J. s ) ( 8.66 \times 10^{13} \ s^{-1})}{480 \ N/m}}[/tex]
[tex]A = \sqrt{\dfrac{5.738116\times 10^{-20} \ m}{480 }[/tex]
[tex]A = \sqrt{1.19544083\times 10^{-22}\ m[/tex]
[tex]A = 1.09336217 \times 10^{-11} \ m[/tex]
[tex]A = 10.9336217 \times 10^{-12} \ m[/tex]
A [tex]\simeq[/tex] 10.9336 pm
