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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.

Electrons are accelerated through a 20 V potential difference producing a monoenergetic beam. This is directed at a double-slit apparatus of 0.010 mm slit separation. A bank of electron detectors is 10 m beyond the double slit. With slit 1 alone open, 100 electrons per second are detected at all detectors. With slit 2 alone open, 900 electrons per second are detected at all detectors. Now both slits are open.

(a) The first minimum in the electron count occurs at detector X. How far is it from the center of the interference pattern?

(b) How many electrons per second will be detected at the center detector?

(c) How many electrons per second will be detected at detector X?

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?

An electron moves along the x-axiswith a well-defined momentum of5脳10-25kg.ms. Write an expression describing the matter wave associated with this electron. Include numerical values where appropriate.

A crack between two walls is 10鈥塩尘wide. What is the angular width of the central diffraction maximum when

(a) an electron moving at 50鈥尘/蝉passes through?

(b) A baseball of mass0.145鈥塳驳 and speed 50鈥尘/蝉passes through?

(c) In each case, an uncertainty in momentum is introduced by the 鈥渆xperiment鈥� (i.e., passing through the slit). Specifically, what aspect of the momentum becomes uncertain, and how does this uncertainty compare with the initial momentum of each?
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