The half-life of (t1/2)an excited state is the time it would take for half the atoms in a large sample to make a transition. Find the relation betweenrole="math" localid="1658300900358" t1/2andT(the “life time” of the state).

Short Answer

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The relation between t1/2andTis thet1/2=τIn2

Step by step solution

01

Half-Life

The half-life of a chemical reaction can be defined as the time taken for the concentration of a given reactant to reach 50% of its initial concentration (I.e., the time taken for the reactant to reach half of its initial value). It is denoted by the symbol t1/2and is usually expressed in seconds.

02

Finding the relation between t1/2 and T (the “life time” of the state)

By definition, the half-life of an excited state is the time it would take for half the atoms to make a transition to a lower state. We use the formula for decay:

Nt=e-tlτN0 (Equation-9.58 and 9.59).

Nbt=Nb0e-At (Equation-9.58).

τ=1A (Equation-9.59).

The half-life of an excited state is the time it would take for half the atoms in a large sample to make the transition.

After one half-life,

Nt=12N0,so12=e-t/τ,or2=et/τ,sot/τ=In2,ort1/2=τIn2

Thus, the relation betweenn role="math" localid="1658302506697" t1/2&Tist1/2=τIn2

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Most popular questions from this chapter

Suppose the perturbation takes the form of a delta function (in time):

H^'=U^δ(t);

Assume thatUaa=Ubb=0,andletUab=Uba+=αif ca(-)=1and cb(-)=0,

find ca(t)andcb(t),and check that lc(t)l2+lcb(t)l2=1. What is the net probability(Pabfort) that a transition occurs? Hint: You might want to treat the delta function as the limit of a sequence of rectangles.

Answer:Pab=sin2(|α|lh)

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|300|nlm|n'l'm'|100.

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Suppose you don’t assumeH'aa=H'aa=0

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Show that

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ϕ(t)1h0l[H'aa(t')-H'bb(t')]dt'.

So the equations fordaanddbare identical in structure to Equation 11.17 (with an extra role="math" localid="1658562498216" eiϕtackedontoH^')

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If the atom is centered at the origin, thenkr1 over the relevant volume,|k|=2π/λ sokr~r/λ1) and that's why we could afford to drop this term. Suppose we keep the first-order correction:

E(r,t)=E0[cos(ωt)+(kr)sin(ωt)].

The first term gives rise to the allowed (electric dipole) transitions we considered in the text; the second leads to so-called forbidden (magnetic dipole and electric quadrupole) transitions (higher powers of k.rlead to even more "forbidden" transitions, associated with higher multipole moments).

(a) Obtain the spontaneous emission rate for forbidden transitions (don't bother to average over polarization and propagation directions, though this should really be done to complete the calculation). Answer:role="math" localid="1659008133999" Rba=q2ω5πϵ0c5|a|(n^r)(k^r)|b|2.

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We have encountered stimulated emission, (stimulated) absorption, and spontaneous emission. How come there is no such thing as spontaneous absorption?

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