# Question Video: Ground State of Quantum Harmonic Oscillator

A quantum mechanical oscillator vibrates at a frequency of 250 THz. What is the least energetic radiation that it can emit?

03:22

### Video Transcript

A quantum mechanical oscillator vibrates at a frequency of 250 terahertz. What is the least energetic radiation that it can emit?

We’re told in the statement that the frequency of oscillation is 250 terahertz; we’ll call that value 𝑓. We want to solve for the least energetic radiation this oscillator can emit, a value that we’ll call 𝐸 sub min. Because we’re working with an oscillator and that the oscillator is quantized, that tells us that we can recall the mathematical relationship describing quantum harmonic oscillator energy levels. The energy level of the 𝑛th state of an oscillator 𝐸 sub 𝑛 is equal to 𝑛 plus one-half quantity times ℎ𝑓, where ℎ is Planck’s constant, which we’ll assume is exactly 6.626 times 10 to the negative 34th joule-seconds.

If we were to sketch out these energy levels, at the ground state, where 𝑛 equals zero, 𝐸 is equal to ℎ𝑓 divided by two. At the first excited state, where 𝑛 equals one, the energy equals three halves ℎ times 𝑓. At the second excited state, where 𝑛 equals two, overall system energy is five halves ℎ times 𝑓. We’re asked to solve for the least energetic radiation that this oscillator can emit. That least energetic radiation would correspond to a step size of one down on the energy levels of the oscillator.

The particular 𝑛 value at which the step starts isn’t important, but what is important is the step size; that’d be one energy level. The step size of one energy level corresponds to a change in energy Δ𝐸 of ℎ times 𝑓 regardless of where that step size occurs. So 𝐸 min is equal to Planck’s constant ℎ times the frequency 𝑓. When we plug in for these two values, making sure to convert our frequency to units of hertz, we can then multiply these two values together. We find then the answer in joules.

But to solve for the result in a unit more fitting for a quantum system, we can find our answer in units of electron volts. To do that, we’ll divide our result in joules by 1.60218 times 10 to the negative 19 joules per electron volt. This conversion factor from joules to electron volts causes the units of joules as a result to cancel out, leaving us with electron volts.

Entering this fraction on our calculator, we find that 𝐸 sub min equals 1.034 electron volts. That is the least energetic radiation this oscillator can emit.