College Physics 11E Global Edition
College Physics 11E Global Edition
11th Edition
ISBN: 9781337620338
Author: SERWAY/VUILLE
Publisher: CENGAGE Learning Custom Publishing
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Chapter 27, Problem 32P

(a)

To determine

The de Broglie wavelength of the electron.

(a)

Expert Solution
Check Mark

Answer to Problem 32P

The de Broglie wavelength of the electron is 5.79×1010m .

Explanation of Solution

The equation for de Broglie wavelength of the electron in terms of kinetic energy is,

λe=h2me(KE)

    • h is the Plank’s constant
    • me is the mass of the electron
    • KE is the kinetic energy

Substitute 6.63×1034J.s for h , 9.11×1031kg for me and 4.50eV for KE .

λe=6.63×1034J.s2(9.11×1031kg)(4.50eV)(1.60×1019J1eV)=5.79×1010m

Conclusion:

Thus, the de Broglie wavelength of the electron is 5.79×1010m .

(b)

To determine

The de Broglie wavelength of the bowling ball.

(b)

Expert Solution
Check Mark

Answer to Problem 32P

The de Broglie wavelength of the bowling ball is 2.26×1025m .

Explanation of Solution

The equation for de Broglie wavelength of the bowling ball in terms of kinetic energy is,

λb=h2mb(KE)

    • mb is the mass of the bowling ball

Substitute 6.63×1034J.s for h , 6.00kg for mb and 4.50eV for KE .

λb=6.63×1034J.s2(6.00kg)(4.50eV)(1.60×1019J1eV)=2.26×1025m

Conclusion:

Thus, the de Broglie wavelength of the bowling ball is 2.26×1025m .

(c)

To determine

The de Broglie wavelength of the photon.

(c)

Expert Solution
Check Mark

Answer to Problem 32P

The de Broglie wavelength of the photon is 2.76×107m .

Explanation of Solution

The equation for de Broglie wavelength of the photon is,

λp=hcE

    • c is the speed of light
    • E is the energy of the photon

Substitute 6.63×1034J.s for h , 3.00×108m/s for c and 4.50eV for E .

λp=(6.63×1034J.s)(3.00×108m/s)(4.50eV)(1.60×1019J1eV)=2.76×107m

Conclusion:

Thus, the de Broglie wavelength of the photon is 2.76×107m

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