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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 9: Q8P (page 356)

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 $5 脳 10^{12} Hz$ , whereas spontaneous emission dominates for frequencies well above . Which mechanism dominates for visible light?

Short Answer

Expert verified

The spontaneous emission dominates.

Step by step solution

01

Formula for the spontaneous emission rate

The spontaneous emission rate is given by:

$A = \frac{蝇^{3} {|鈩�|}^{2}}{3 {蟺辞}_{0} h c^{3}}$

The thermally stimulated emission rate is given by:

role="math" localid="1658378927515" $R = \frac{蟺}{3 o_{0} h^{2}} {|鈩�|}^{2} 蚁 (蝇)$

where:

$蚁 (蝇) = \frac{h}{蟺^{2} c^{3}} \frac{蝇^{3}}{(e^{h 蝇 / k_{B} T} - 1)}$

The ratio of the spontaneous emission rate to the thermally stimulated emission rate is,

$\frac{A}{R} = \frac{蝇^{3} {|鈩�|}^{2}}{3 {蟺辞}_{0} h c^{3}} . \frac{3 {蟺辞}_{0} h^{2}}{蟺 {|鈩�|}^{2}} . \frac{蟺^{2} c^{3} (e^{h 蝇 / K_{B} T} - 1)}{h 蝇^{3}} = e^{h 蝇 / K_{B} T} - 1$

02

Find out the result

Seek the point at which the ratio is, that is:

$1 = e^{h 蝇 / K_{B} T} - 1 e^{h 蝇 / K_{B} T} - 2 \frac{h 蝇}{k_{B} T} = I n 2 蝇 = \frac{k_{B} T}{h} I n 2$

The frequency is,

$v = \frac{蝇}{2 蟺} = \frac{k_{B} T}{h} I n 2$

At room temperature, we get:

$v = \frac{(1.38 脳 10^{- 23} J / k) (300 k)}{6.63 脳 10^{- 34} J . s} = 4.33 脳 10^{12} Hz$

For a frequency higher than this frequency the spontaneous emission dominates, note that higher frequencies than this one includes the visible light.

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Most popular questions from this chapter

A hydrogen atom is placed in a (time-dependent) electric $field E = E (t) \overset{铃�}{k .} calculate all four matrix elements H_{ij}^{,} of the perturbation \overset{铃�}{H^{,}} = eEz$ between the ground state (n = 1 ) the (quadruply degenerate) first excited states (n = 2 ) . Also show ${thatH}_{ii}^{,} = 0$ 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 鈥渁ccessible鈥� from the ground state by a perturbation of this form, and therefore the system functions as a two-state configuration鈥攁ssuming transitions to higher excited states can be ignored.

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

Calculate $c_{a} (t) and c_{b} (t)$ , to second order, for a time-independent perturbation in Problem 9.2. Compare your answer with the exact result.

Close the 鈥渓oophole鈥� in Equation 9.78 by showing that if $l^{'} = l = 0$ then $鉄� n^{'} l^{'} m^{'} | r | n l m 鉄� = 0$

A particle starts out (at time t=0 ) in the Nth state of the infinite square well. Now the 鈥渇loor鈥� of the well rises temporarily (maybe water leaks in, and then drains out again), so that the potential inside is uniform but time dependent: $V_{0} (t), with V_{0} (0) = V_{0} (T) = 0$ .

(a) Solve for the exact $c_{m} (t)$ , using Equation 11.116, and show that the wave function changes phase, but no transitions occur. Find the phase change, role="math" localid="1658378247097" $蠒 (T)$ , in terms of the function $V_{0} (t)$

(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.

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