Question

Louis de Broglie's bold hypothesis assumes that it is possible to assign a wavelength λ to every particle possessing some momentum p by the relationship λ = h p , where h is Planck's constant ( h = 6.626 × 10 − 34 J ⋅ s ). This applies not only to subatomic particles like electrons, but every particle and object that has a momentum. To help you develop some number sense for de Broglie wavelengths of common, everyday objects, try below calculations. Use Planck's constant h = 6.626 × 10 − 34 J ⋅ s ; other necessary constants will be given below. To enter answers in scientific notation below, use the exponential notation. For example, 3.14 × 10 − 14 would be entered as "3.14E-14".

Air molecules (mostly oxygen and nitrogen) move at speeds of about 270 m/s. If mass of air molecules are about 5 × 10 − 26 kg , their de Broglie wavelength is m. Consider a baseball thrown at speed 50 m/s. If mass of the baseball is 0.14 kg, its de Broglie wavelength is m. The Earth orbits the Sun at a speed of 29800 m/s. Given that the mass of the Earth is about 6.0 × 10 24 kg , its de Broglie wavelength is m. Yes, many of these numbers are absurdly small, which is why I think you should enter the powers of 10.

Answer #1

(i) Momentum of air molecules is given by -

p_{air} = m_{air} v = [(5 x 10^{-26} kg)
(270 m/s)]

p_{air} = 1.35 x 10^{-23} kg.m/s

Their de Broglie wavelength of air which will be given by -

_{air} = h
p_{air} = [(6.626 x 10^{-34} J.s) (1.35 x
10^{-23} kg.m/s)]

_{air} = 8.94 x
10^{-57} m

Answer in scientific notation : "8.94E-57"

(ii) Momentum of a baseball is given by -

p_{ball} = m_{ball} v = [(0.14 kg) (50 m/s)]

p_{ball} = 7 kg.m/s

Their de Broglie wavelength of baseball which will be given by -

_{ball} = h
p_{ball} = [(6.626 x 10^{-34} J.s) (7 kg.m/s)]

_{ball} = 4.63 x
10^{-33} m

Answer in scientific notation : "4.63E-33"

(iii) Momentum of the Earth is given by -

p_{Earth} = m_{Earth} v = [(6 x 10^{24}
kg) (29800 m/s)]

p_{Earth} = 1.78 x 10^{29} kg.m/s

Their de Broglie wavelength of air which will be given by -

_{Earth} = h
p_{Earth} = [(6.626 x 10^{-34} J.s) (1.78 x
10^{29} kg.m/s)]

_{Earth} = 1.17 x
10^{-4} m

Answer in scientific notation : "1.17E-4"

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