It is also possible to express this in terms of uncertainty in energy and in tim

Question

It is also possible to express this in terms of uncertainty in energy and in time. This formulation of the HUP states delta E delta t greaterthanorequalto h/2. One implication of this is that the precision with which the energy of some spectroscopic absorption can be measured depends on the time duration of the process. For instance, elements present in distant nebulae and galaxies can be identified based on their atomic emission spectra. Not only do atoms emit at distinct wavelengths, as addressed in week 2 of the semester, but they also have excited states of specific lifetimes. Calculate the wavelength of light expected to be emitted as atomic hydrogen relaxes from its first excited state to its ground state. Assuming this excited state has a lifetime of 1.6 ns. what is the absolute uncertainty in energy of the photon you expect to be emitted? What is the relative uncertainty in energy? What range of wavelengths would this represent?

Explanation / Answer

part a:

maximum energy of the photon=(13.6 eV)*((1/n1^2)-(1/n2^2))

where n1=ground state index=1

n2=first excited state index=2

then maximum energy=13.6*(1-0.25)=10.2 eV

then if wavelength of released photon is lambda,

h*c/lambda=10.2 eV=10.2*1.6*10^(-19) J

where h=planck's constant=6.626*10^(-34)

and c=speed of light=3*10^8

==>lambda=h*c/energy

=6.626*10^(-34)*3*10^8/(10.2*1.6*10^(-19))=1.218*10^(-7) m=121.8 nm

part b:

uncertainty in time=1.6 ns=1.6*10^(-9) seconds

then uncertainty in energy=(h/(4*pi))/uncertainty in time

=6.626*10^(-34)/(4*pi*1.6*10^(-9))=3.2955*10^(-26) J

part c:

relative uncertainty=uncertainty/energy value

=3.2955*10^(-26)/(10.2*1.6*10^(-19))=2.0193*10^(-8)

part d:

lambda=h*c/energy

change in lambda=(-h*c/energy^2)*change in energy

==(-6.626*10^(-34)*3*10^8/(10.2*1.6*10^(-19))^2)*3.2955*10^(-26)

=-2.4595*10^(-15) m

so range of wavelength is 2.4595*10^(-15) m .

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