# 2.1

Calculate the de Broglie wavelength and the kinetic energy of each of the following:

1. A 10 kg toddler jogging on a treadmill with a velocity of 1 m/s
2. A 1 g silver bullet fired at a velocity of 10,000 m/s
3. The moon as it orbits the Earth
4. A positron with a velocity of $$10 \times 10^6$$ cm/s
5. A chicken crossing the road with a velocity of 1 cm/s. Assume the chicken has a mass of 1 kg.
6. An electron in the n = 1 Bohr orbital which has a velocity of 2.19 x 10m/s

## Text Question

Calculate the de Broglie wavelength and the kinetic energy of each of the following:

1. A 10 kg toddler jogging on a treadmill with a velocity of 1 m/s
2. A 1 g silver bullet fired at a velocity of 10,000 m/s
3. The moon as it orbits the Earth
4. A positron with a velocity of $$10 \times 10^6$$ cm/s
5. A chicken crossing the road with a velocity of 1 cm/s. Assume the chicken has a mass of 1 kg.
6. An electron in the n = 1 Bohr orbital which has a velocity of 2.19 x 10m/s

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## Solution

All of these parts are simple algebra calculations with the de Broglie equation.

$\lambda = \dfrac{h}{p} = \dfrac{h}{mv} \nonumber$

a. A 10 kg toddler jogging on a treadmill with a velocity of 1 m/s

$\lambda = \dfrac{h}{10 \;kg \times 1 \;m/s} = 6.626 \times 10^{-35} \;m \nonumber$

$KE = \dfrac{(10 \;kg)(1 \;m/s)^{2}}{2} = 5 \;J\nonumber$

b. A 1 g silver bullet fired at a velocity of 10,000 m/s

$\lambda = \dfrac{h}{1 \;g \times 10000 \;m/s} = 6.626 \times 10^{-35} \;m\nonumber$

$KE = \dfrac{(.001 \;kg)(10000 \;m/s)^{2}}{2} = 50,000 \;J\nonumber$

c. The moon as it orbits the Earth

From Wikipedia $$m_{moon} = 7.3 \times 10^{22} \;kg$$ and $$v_{orbital} = 1022 \;m/s$$

$\lambda = \dfrac{h}{7.3 \times 10^{22} \;kg \times 1022 \;m/s} = 8.881 \times 10^{-60} \;m\nonumber$

$KE = \dfrac{(7.3 \times 10^{22} \;kg)(1022 \;m/s)^{2}}{2} = 3.8 \times 10^{28} \;J\nonumber$

d. A positron with a velocity of $$10 \times 10^6$$ cm/s

A positron is an electron with a positive charge (a simplification of anti-matter)

$\lambda = \dfrac{h}{9.1 \times 10^{-31} \;kg \times 10 \times 10^4 \;m/s} = 7.3 \times 10^{-9} \;m = 7.3 \;nm\nonumber$

$KE = \dfrac{(9.1 \times 10^{-31} \;kg)(10 \times 10^{4} \;m/s)^{2}}{2} = 4.6 \times 10^{-21} \;J\nonumber$

The first result that has a meaningful wavelength.

e. A chicken crossing the road with a velocity of 1 cm/s. Assume the chicken has a mass of 1 kg.

$\lambda = \dfrac{h}{1 \;kg \times 0.01 \;m/s} = 6.626 \times 10^{-32} \;m\nonumber$

$KE = \dfrac{(1 \;kg)(.01 \;m/s)^{2}}{2} = 5 \times 10^{-5} \;J\nonumber$

f. An electron in the n = 1 Bohr orbital which has a velocity of $$2.19 \times 10^6$$ m/s

The $$n=1$$ Bohr orbital is extra information we don't need.

$\lambda = \dfrac{h}{9.1 \times 10^{-31} \;kg \times 2.19 \times 10^6 \;m/s} =3.32 \times 10^{-10} \;m = 0.332 nm \nonumber$

$KE = \dfrac{(9.1 \times 10^{-31} \;kg)(2.19 \times 10^{6} \; m/s)^{2}}{2} = 2.18 \times 10^{-18} \;J\nonumber$

In all these deBroglie wavelength questions pay attention to when the wavelength actually matters. Only when the object is very small with a mass of about the same order $$10^{-30}$$ ish as $$h$$ is the wavelength important. Quantum effects became too small to ever notice with everyday "real" objects.

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