Transition between excited states

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
8 replies · 6K views
gildomar
Messages
98
Reaction score
2

Homework Statement



An atom in an excited state has a lifetime of 1.2 x 10 -8 sec; in a second excited state the
lifetime is 2.3 x 10 -8 sec. What is the uncertainty in energy for the photon emitted when
an electron makes a transition between these two levels?

Homework Equations



[itex]\Delta[/itex]E[itex]\Delta[/itex]t[itex]\geq[/itex][itex]\frac{\hbar}{2}[/itex]

The Attempt at a Solution



So I just found the uncertainty in energies for the two excited states using the uncertainty principle, getting 2.74*10^-8 eV for the 1.2*10^-8 sec state, and 1.43*10^-8 eV for the 2.3*10^-8 sec state. And figured that the uncertainty in energy would just be the difference in the energies, giving 1.31*10^-8 eV. But the book gives an answer of 4.17*10^-8 eV, which I noticed is what you get if you add the energies instead.

So is the book wrong, or is there some weird thing about the uncertainties combining such that I have to add them instead?
 
Physics news on Phys.org
Shouldn't it follow from

[itex]\Delta E \Delta t \geq \frac{\hbar}{2}[/itex]

that

[itex]\Delta E \geq \frac{\hbar}{2 \Delta t}[/itex]

?
 
Whether adding or subtracting two quantities, the uncertainties add.
 
@Basic_Physics: Yes, that's how I got the energies.

@gneill: That's a rule from statistics? Cause the book was a bit sparse on that point.
 
gildomar said:
@gneill: That's a rule from statistics? Cause the book was a bit sparse on that point.
Yup. Consider that subtraction is just adding the negative of one of the values. The uncertainty in the negative value is the same as for the positive value. So in terms of uncertainty, addition and subtraction are the same.
 
Thanks gneill! At least now I know I'm not going crazy.
 
gildomar said:
Thanks gneill! At least now I know I'm not going crazy
I'm kinda confused here, what have you resolved at?? Is your value correct or wrong ??
 
A good way to realize that you don't subtract the uncertainties is to imagine that the two uncertainies happen to be equal.

Subtracting would give a result of zero uncertainty, which (I hope you see) makes no sense!