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## Homework Statement

According to the energy-time uncertainty principle, the lifetime t of a state is inversely proportional to the uncertainty in the energy E.

We consider the line λ= 656nm resulting from a transition in a hydrogen atom,

from an excited state of lifetime 10

^{-8}s.

(a) What is the uncertainty in the energy of the emitted photon?

(b) What is the corresponding uncertainty in the wavelength, Δλ?

My question involves b, but let's do a first.

## Homework Equations

[tex]\Delta E \Delta t \approx \hbar[/tex]

## The Attempt at a Solution

So for a I just solved the above equation for ΔE which gave me.

[tex]\Delta E \approx \frac{\hbar}{\Delta t}[/tex]

Now part a is done. (the value doesn't matter for my actual question)

Next up is b, where my question comes in.

This is how I solved it, which gave me a huge wavelength (in meters) and was wrong.

I had....

[tex]E=hf = \frac{hc}{\lambda}[/tex]

Solving for lambda gave:

[tex]\lambda=\frac{hc}{E}[/tex]

I actually had it right up to this point.

The correct method was to take the derivative of what I had for E with respect to λ.

Which gives:

[tex]\frac{dE}{d\lambda}=-\frac{hc}{\lambda^2}[/tex]

Or something like that. I took notes on the paper I handed in.

The actual answer after the derivative was...which I don't really get either. I'm not sure how we have λ

^{2}in the denominator and then the square goes away and a Δλ jumps up to the top.

[tex]E=\left|-\frac{hc}{\lambda}\Delta \lambda\right|[/tex]

This is then obviously solved for Δλ.

Finally...the question.

Why does one take the derivative in order to get the uncertainty for E? We didn't set it to zero or anything. So I don't think it was min/max type stuff.

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