Question

The effective spring constant describing the potential energy of the HBr molecule is 410 N/m and that for the NO molecule is 1530 N/m.A) Calculate the minimum amplitude of vibration for the NO molecule.
B) Calculate the minimum amplitude of vibration for the HCl molecule.

Answer 1

Answer:

a. the minimum amplitude of vibration for the NO molecule A $\simeq$ 4.9378 pm

b. the minimum amplitude of vibration for the HCl molecule A $\simeq$ 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:

$\dfrac{1}{2}kA^2= \dfrac{1}{2}hf$

$kA^2= hf$

$A^2= \dfrac{hf}{k}$

$A = \sqrt{\dfrac{hf}{k}}$

$A = \sqrt{\dfrac{(6.626 \times 10^{-34} \ J. s ) ( 5.63 \times 10^{13} \ s^{-1})}{1530 \ N/m}}$

$A = \sqrt{\dfrac{3.730438 \times 10^{-20} \ m}{1530 }$

$A = \sqrt{2.43819477 \times 10^{-23}\ m$

$A =4.93780799 \times 10^{-12} \ m$

A $\simeq$ 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:

$A = \sqrt{\dfrac{hf}{k}}$

$A = \sqrt{\dfrac{(6.626 \times 10^{-34} \ J. s ) ( 8.66 \times 10^{13} \ s^{-1})}{480 \ N/m}}$

$A = \sqrt{\dfrac{5.738116\times 10^{-20} \ m}{480 }$

$A = \sqrt{1.19544083\times 10^{-22}\ m$

$A = 1.09336217 \times 10^{-11} \ m$

$A = 10.9336217 \times 10^{-12} \ m$

A $\simeq$ 10.9336 pm