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

A particle starts out (at time t=0 ) in the Nth state of the infinite square well. Now the “floor” of the well rises temporarily (maybe water leaks in, and then drains out again), so that the potential inside is uniform but time dependent:V0(t),withV0(0)=V0(T)=0.

(a) Solve for the exact cm(t), using Equation 11.116, and show that the wave function changes phase, but no transitions occur. Find the phase change, role="math" localid="1658378247097" ϕ(T), in terms of the function V0(t)

(b) Analyze the same problem in first-order perturbation theory, and compare your answers. Compare your answers.
Comment: The same result holds whenever the perturbation simply adds a constant (constant in x, that is, not in to the potential; it has nothing to do with the infinite square well, as such. Compare Problem 1.8.

Calculate ca(t)andcb(t), to second order, for a time-independent perturbation in Problem 9.2. Compare your answer with the exact result.

A hydrogen atom is placed in a (time-dependent) electric fieldE=E(t)k.calculateallfourmatrixelementsHij,oftheperturbationH,=eEzbetween the ground state (n = 1 ) the (quadruply degenerate) first excited states (n = 2 ) . Also showthatHii,=0 for all five states. Note: There is only one integral to be done here, if you exploit oddness with respect to z; only one of the n = 2 states is “accessible” from the ground state by a perturbation of this form, and therefore the system functions as a two-state configuration—assuming transitions to higher excited states can be ignored.

As a mechanism for downward transitions, spontaneous emission competes with thermally stimulated emission (stimulated emission for which blackbody radiation is the source). Show that at room temperature (T = 300 K) thermal stimulation dominates for frequencies well below 5×1012Hz , whereas spontaneous emission dominates for frequencies well above . Which mechanism dominates for visible light?

Prove the commutation relation in Equation 9.74. Hint: First show that

[L2,z]=2ih(xLy-yLx-ihz)

Use this, and the fact that localid="1657963185161" r.L=r.(r×p)=0, to demonstrate that

[L2,[L2,z]]=2h2(zL2+L2z)

The generalization from z to r is trivial.

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