2.1
- Page ID
- 102688
Calculate the de Broglie wavelength and the kinetic energy of each of the following:
- A 10 kg toddler jogging on a treadmill with a velocity of 1 m/s
- A 1 g silver bullet fired at a velocity of 10,000 m/s
- The moon as it orbits the Earth
- A positron with a velocity of \(10 \times 10^6\) cm/s
- A chicken crossing the road with a velocity of 1 cm/s. Assume the chicken has a mass of 1 kg.
- An electron in the n = 1 Bohr orbital which has a velocity of 2.19 x 106 m/s
Text Question
Calculate the de Broglie wavelength and the kinetic energy of each of the following:
- A 10 kg toddler jogging on a treadmill with a velocity of 1 m/s
- A 1 g silver bullet fired at a velocity of 10,000 m/s
- The moon as it orbits the Earth
- A positron with a velocity of \(10 \times 10^6\) cm/s
- A chicken crossing the road with a velocity of 1 cm/s. Assume the chicken has a mass of 1 kg.
- An electron in the n = 1 Bohr orbital which has a velocity of 2.19 x 106 m/s
A11YQuestion
N/A
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.
Hint
N/A
Link
N/A
Notes
N/A