# Calculating Spring Constant using Minimum Energy State of hydrogen atom

• calphyzics09
N/m}In summary, to find the "spring constant" for a hydrogen chloride molecule, we can use the energy eigenvalues of the quantum harmonic oscillator. By setting n=1 for the first excited state and converting the minimum photon energy (0.358 eV) to J, we can calculate the spring constant to be approximately 220 N/m.
calphyzics09

## Homework Statement

A hydrogen chloride molecule may be modeled as a hydrogen atom (mass: 1.67 x10^-27 kg ) on a spring; the other end of the spring is attached to a rigid wall (the massive chlorine atom).

If the minimum photon energy that will promote this molecule to its first excited state is 0.358 eV, find the "spring constant."

I'm not sure which equation to use. is it E=1/2kA^2? If so, what would the value of A be?

Thank you for your help!

Shouldn't you be using the energy eigenvalues of the quantum harmonic oscillator, and not the classical potential energy of a spring?

I see...so what does that mean? I'm kind of confused

calphyzics09 said:
I see...so what does that mean? I'm kind of confused

I assume by this comment that you haven't covered the http://galileo.phys.virginia.edu/classes/252/SHO/SHO.html" . I'd recommend taking a glimpse over that link, it's a pretty simplistic essay on the quantum harmonic oscillator. But basically the energy eigenvalues of the QHO are given by

$$E_n=\hbar\omega\left(n+\frac{1}{2}\right)$$

where $\omega^2=k/m$. So in the ground state, your energy equation would be

$$E_0=\frac{1}{2}\hbar\sqrt{\frac{k}{m}}$$

So you can use this equation to solve for your spring constant $k$. (Just in the off chance you don't know what it is, $\hbar=1.054\times10^{-34}\,\mathrm{m^2kg/s}$ and is the reduced Planck constant)

Edit: that was silly of me, forgot the power in $\hbar$!

Last edited by a moderator:
hmmm I used that formula and got 7.7 x 10^40..which is incorrect..am I using the wrong units? Thanks for your help btw

You need to do two things:
(1) convert eV into J: $0.358 \,\mathrm{eV}=5.73\times10^{-20}\mathrm{m^2kg/s^2}$.
(2) let $n=1$ so that you can have the first excited state. I didn't catch this one earlier, but your energy eigenvalue should be

$$E_1=\frac{3\hbar}{2}\sqrt{\frac{k}{m}}$$

leading to

$$k=\frac{4mE_1^2}{9\hbar^2}=\frac{4\cdot1.67 \times10^{-27} \mathrm{kg}\cdot(5.73\times10^{-20}\mathrm{m^2kg/s^2})^2}{9(1.05\times10^{-34}\mathrm{m^2kg/s^2})^2}\approx220$$

## 1. What is the minimum energy state of a hydrogen atom?

The minimum energy state of a hydrogen atom is the ground state, where the electron is in the lowest energy level and closest to the nucleus.

## 2. How is the spring constant of a hydrogen atom calculated using its minimum energy state?

The spring constant of a hydrogen atom can be calculated by using the formula k = (e^2)/(4πε_0a_0^3), where e is the electron charge, ε_0 is the permittivity of free space, and a_0 is the Bohr radius.

## 3. What is the significance of calculating the spring constant of a hydrogen atom?

Calculating the spring constant of a hydrogen atom allows us to understand the strength of the interaction between the electron and the nucleus, and how it affects the atom's energy levels and behavior.

## 4. Can the spring constant of a hydrogen atom be changed?

The spring constant of a hydrogen atom is a constant value that depends on fundamental physical constants and cannot be changed unless these constants are altered.

## 5. How does the calculated spring constant of a hydrogen atom compare to experimental values?

The calculated spring constant of a hydrogen atom is in good agreement with experimental values, providing further evidence for the validity of the model used to describe the atom's behavior.

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