Question

# Louis de Broglie's bold hypothesis assumes that it is possible to assign a wavelength λ to...

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.

(i) Momentum of air molecules is given by -

pair = mair v = [(5 x 10-26 kg) (270 m/s)]

pair = 1.35 x 10-23 kg.m/s

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

air = h pair = [(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 -

pball = mball v = [(0.14 kg) (50 m/s)]

pball = 7 kg.m/s

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

ball = h pball = [(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 -

pEarth = mEarth v = [(6 x 1024 kg) (29800 m/s)]

pEarth = 1.78 x 1029 kg.m/s

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

Earth = h pEarth = [(6.626 x 10-34 J.s) (1.78 x 1029 kg.m/s)]

Earth = 1.17 x 10-4 m

Answer in scientific notation : "1.17E-4"

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