Determining the life time of a excited state.

[Elekko]

Homework Statement

The smallest uncertainty in the frequency of emitted light when excited atoms return to ground state for molecules is estimated to be 8 MHz. Use this information to estimate the lifetime of the excited states.
I would like to know if I'm thinking correct.

Homework Equations

Well here, it is a question we can take in general by having an uncertainty of the frequency at f = MHz.
I took the hydrogen atom in which the energy levels in general can be written as

$E_n=\frac{-13.6eV}{n^2}$ where I then calculate the difference in energy for instance between state n = 1 and n = 2.

The Attempt at a Solution

Can I then apply the energy-time uncertainty principle $\Delta E \Delta t \ge \frac{\hbar}{2}$ ?
I'm not sure about this, science we have an uncertainty in FREQUENCY which makes me stuck.

Appreciate for help.

 

[TSny]

Recall the general relation E = hf.

 

[Elekko]

Recall the general relation E = hf.

Can this be used as the uncertainty in energy in the energy-time uncertainty principle?
(Im not very sure about this, science I don't have a solution for this)

 

[Elekko]

Need to mention that of course f = 8 MHz

 

[paranormal]

Yes, you are on the right track. The energy-time uncertainty principle states that the uncertainty in energy (\Delta E) multiplied by the uncertainty in time (\Delta t) must be greater than or equal to hbar/2, where hbar is the reduced Planck's constant. In this case, the uncertainty in energy is related to the frequency by the equation E = hf, where h is Planck's constant and f is the frequency. So, we can rewrite the energy-time uncertainty principle as \Delta E \Delta t \geq \frac{\hbar}{2} \implies \Delta f \Delta t \geq \frac{1}{4\pi}.

In order to determine the lifetime of the excited state, we need to first calculate the uncertainty in time. We can do this by rearranging the equation for energy levels of the hydrogen atom to solve for the energy difference between states n = 1 and n = 2:

E_2 - E_1 = \frac{-13.6eV}{2^2} - \frac{-13.6eV}{1^2} = 3.4eV

Now, we can use the equation E = hf to calculate the uncertainty in frequency:

\Delta f = \frac{\Delta E}{h} = \frac{3.4eV}{h} = \frac{3.4eV}{6.63 \times 10^{-34} J \cdot s} = 5.13 \times 10^{14} Hz

Finally, we can use the uncertainty in frequency and the given uncertainty in frequency (8 MHz) to calculate the uncertainty in time:

\Delta f \Delta t \geq \frac{1}{4\pi} \implies (5.13 \times 10^{14} Hz)(\Delta t) \geq \frac{1}{4\pi} \implies \Delta t \geq \frac{1}{4\pi(5.13 \times 10^{14} Hz)} = 3.07 \times 10^{-17} s

Therefore, the estimated lifetime of the excited state is 3.07 x 10^-17 seconds. Keep in mind that this is just an estimate and the actual lifetime may vary.