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Chapter 38: Q 67P (page 1184)

Question:The uncertainty in the position of an electron along an xaxis

is given as 50pm, which is about equal to the radius of a hydrogen

atom. What is the least uncertainty in any simultaneous measurement

of the momentum component of this electron?

Short Answer

Expert verified

The least uncertainty in the momentum component pxof this electron is

∆px=2.1×10-24kgms

Step by step solution

01

Identifying the data given in the question.

The uncertainty in the position of the electron is

∆x=50pm=50×10-12m

02

Concept used to solve the question

Heisenberg’s Uncertainty Principle

According to Heisenberg’s Uncertainty Principle, the uncertainty in position and momentum are related as,

∆x∆px≥h2π

Where;

∆x is uncertainty in position along the axis.

∆pxis uncertainty in momentum along the axis.


his plank’s constant.

03

Finding the least uncertainty in the momentum component px

According to Heisenberg’s Uncertainty Principle

∆x∆px≥h2π

The least uncertainty possible when

∆x∆px=h2π

Therefore,

∆px=h2π∆x

Substituting the values,

∆px=6.634×10-34J.s2π50×10-12m=2.1×10-24kgms-1

Hence the least uncertainty in the momentum component pxof this electron is

∆px=2.1×10-24kgms-1

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Most popular questions from this chapter

The highest achievable resolving power of a microscope is limited only by the wavelength used; that is, the smallest item that can be distinguished has dimensions about equal to the wavelength. Suppose one wishes to “see” inside an atom. Assuming the atom to have a diameter of 100pm, this means that one must be able to resolve a width of, say, 10pm.

(a) If an electron microscope is used, what minimum photon energy is required?

(b) If a light microscope is used, what minimum photon energy is required?

(c) Which microscope seems more practical? Why?

Let K be the kinetic energy that a stationary free electron gains when a photon scatters from it. We can plot K versus the angle ϕat which the photon scatters; see curve 1 in Fig. 38-21. If we switch the target to be a stationary free proton, does the end point of the graph shift (a) upward as suggested by curve 2, (b) downward as suggested by curve 3, or (c) remain the same?

Assuming that your surface temperature is98.6° Fand that you are an ideal blackbody radiator (you are close), find

(a) the wavelength at which your spectral radiancy is maximum,

(b) the power at which you emit thermal radiation in a wavelength range of 1.00 nmat that wavelength, from a surface area of4.00 cm2, and

(c) the corresponding rate at which you emit photons from that area. Using a wavelength of500 nm (in the visible range),

(d) recalculate the power and

(e) the rate of photon emission. (As you have noticed, you do not visibly glow in the dark.)

Question:Consider a potential energy barrier like that of Fig. 38-17but whose height Ubis and 6eVwhose thickness Lis 0.70nm. What is the energy of an incident electron whose transmissioncoefficient is 0.0010?

Calculate the percentage change in photon energy during collision like that in Fig. 38-5 forϕ=90∘ and for radiation in

(a) the microwave range, withλ=3.0 cm ;

(b) the visible range, with λ=500 nm;

(c) the x-ray range, withλ=25 pm ; and

(d) the gamma-ray range, with a gamma photon energy of 1.0 MeV.

(e) What are your conclusions about the feasibility of detecting the Compton shift in these various regions of the electromagnetic spectrum, judging solely by the criterion of energy loss in a single photon-electron encounter?
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