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Chapter 4: Q47E (page 137)

An electron in an atom can "jump down" from a higher energy level to a lower one, then to a lower one still. The energy the atom thus loses at each jump goes to a photon. Typically, an electron might occupy a level for a nanosecond. What uncertainty in the electron's energy does this imply?

Short Answer

Expert verified

The uncertainty in the electron鈥檚 energy is 鈭�E鈮�5.3脳10-23J.

Step by step solution

01

Given data

Time is given as: 鈭�t=10-9s.

02

Uncertainty principle

The principle states that the position and the velocity of an object cannot be measured with 100 % accuracy at the same time.

鈭�x路鈭�p鈮�h2

鈭�x= Uncertainty in the position.

鈭�p = Uncertainty of momentum.

h = Planck's constant. = 1.05x10-34 J.s

03

Electron’s energy

The Energy and Time Uncertainty Principle,

鈭�t路鈭�E鈮�h2

Substituting values, and we get:

role="math" localid="1658386938701" 鈭�E鈮�1.05脳10-342脳1脳10-9鈭�E鈮�5.3脳10-26J

Therefore, the total energy is 鈭�E鈮�5.3脳10-26J.

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Most popular questions from this chapter

One of the cornerstones of quantum mechanics is that bound particles cannot be stationary-even at zero absolute temperature! A "bound" particle is one that is confined in some finite region of space. as is an atom in a solid. There is a nonzero lower limit on the kinetic energy of such a particle. Suppose minimum kinetic energy of width L. Obtain an approximate formula for its minimum kinetic energy.

In Exercise 45, the case is made that the position uncertainty for a typical macroscopic object is generally so much smaller than its actual physical dimensions that applying the uncertainty principle would be absurd. Here we gain same idea of how small an object would have鈾� to be before quantum mechanics might rear its head. The density of aluminum is 2.7脳103鈥�kg/m3, is typical of solids and liquids around us. Suppose we could narrow down the velocity of an aluminum sphere to within an uncertainty of1渭mper decade. How small would it have to be for its position uncertainty to be at least as large as110%of its radius?

Question: Atoms in a crystal form atomic planes at many different angles with respect to the surface. The accompanying figure shows the behaviors of representative incident and scattered waves in the Davisson-Germer experiment. A beam of electrons accelerated through 54 V is directed normally at a nickel surface, and strong reflection is detected only at an angle 蠒of 500.Using the Bragg law, show that this implies a spacing D of nickel atoms on the surface in agreement with the known value of 0.22 nm.

The Moon orbits Earth at a radius of 3.84脳108m. To do so as a classical particle. Its wavelength should be small. But small relative to what? Being a rough measure of the region where it is confined, the orbit radius is certainly a relevant dimension against which to compare the wavelength. Compare the two. Does the Moon indeed orbit as a classical particle? (localid="1659095974931" mEarth=5.98脳1024kgand mmoon=7.35脳1022kg)

Because we have found no way to formulate quantum mechanics based on a single real wave function, we have a choice to make. In Section 4.3,it is said that our choice of using complex numbers is a conventional one. Show that the free-particle Schrodinger equation (4.8) is equivalent to two real equations involving two real functions, as follows:

-鈩�2位2唯1(x,t)鈭�m=鈩�鈭�唯2(x,t)鈭�tand

-鈩�2位2唯2(x,t)鈭�m=鈩�鈭�唯1(x,t)鈭�t

where 唯(x,t)is by definition 唯1(x,t)+i唯2(x,t). How is the complex approach chosen in Section4.3more convenient than the alternative posed here?
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