College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Chapter 27, Problem 46AP

(a)

To determine

Johnny’s de Broglie wavelength at the moment.

(a)

Expert Solution
Check Mark

Answer to Problem 46AP

Johnny’s de Broglie wavelength at the moment is 2.82×1037m .

Explanation of Solution

Johnny’s speed just before impact is,

v2=v02+2ay(Δy)v=v02+2ay(Δy)

    • v0 is the initial velocity
    • ay is the vertical acceleration
    • Δy is the vertical displacement

Substitute 0 for v0 , 9.80m/s for ay and 50.0m for Δy .

v=0+2(9.80m/s)(50.0m)=31.3m/s

The equation for de Broglie wavelength is,

λ=hmv

    • h is the Plank’s constant
    • m is the mass
    • v is the velocity

Substitute 6.63×1034J.s for h , 75.0kg for m and 31.3m/s for v .

λ=6.63×1034J.s(75.0kg)(31.3m/s)=2.82×1037m

Conclusion:

Thus, Johnny’s de Broglie wavelength at the moment is 2.82×1037m .

(b)

To determine

The uncertainty of the kinetic energy measurement.

(b)

Expert Solution
Check Mark

Answer to Problem 46AP

The uncertainty of the kinetic energy measurement is 1.06×1032J .

Explanation of Solution

The equation for energy uncertainty is,

ΔEh4π(Δt)

Substitute 6.63×1034J.s for h and 5.00ms for Δt .

ΔE=6.63×1034J.s4π(5.00ms)(103s1ms)=1.06×1032J

Conclusion:

Thus, the uncertainty of the kinetic energy measurement is 1.06×1032J .

(c)

To determine

The percentage error caused by the uncertainty.

(c)

Expert Solution
Check Mark

Answer to Problem 46AP

The percentage error caused by the uncertainty is 2.88×1035% .

Explanation of Solution

The percentage error is,

%error=ΔEmg|Δy|(100%)

Substitute 1.06×1032J for ΔE , 75.0kg for m , 9.80m/s2 for g and 50.0m for Δy .

%error=1.06×1032J(75.0kg)(9.80m/s2)(50.0m)(100%)=2.88×1035%

Conclusion:

Thus, the percentage error caused by the uncertainty is 2.88×1035%

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