Which of the following has the smallest de Broglie wavelength? a) an electron traveling at \(80 \%\) the speed of light b) a proton traveling at \(20 \%\) the speed of light c) a carbon nucleus traveling at \(70 \%\) the speed of light d) a helium nucleus traveling at \(80 \%\) the speed of light e) a lithium nucleus traveling at \(50 \%\) the speed of light

Short Answer

Expert verified
a) an electron at \(80\%\) the speed of light b) a proton at \(20\%\) the speed of light c) a carbon nucleus at \(70\%\) the speed of light d) a helium nucleus at \(80\%\) the speed of light e) a lithium nucleus at \(50\%\) the speed of light Answer: c) a carbon nucleus at \(70\%\) the speed of light

Step by step solution

01

Write down de Broglie wavelength formula

\(\lambda = \frac{h}{mv}\)
02

Find the mass of each particle

Mass of an electron: \(m_e = 9.11 \times 10^{-31} kg\) (approx.) Mass of a proton: \(m_p = 1.67 \times 10^{-27} kg\) (approx.) Mass of a carbon nucleus: \(m_C = 12 \times m_p\) Mass of a helium nucleus: \(m_{He} = 4 \times m_p\) Mass of a lithium nucleus: \(m_{Li} = 7 \times m_p\)
03

Calculate the velocity of each particle

Velocity of an electron: \(v_e = 0.8c\) Velocity of a proton: \(v_p = 0.2c\) Velocity of a carbon nucleus: \(v_C = 0.7c\) Velocity of a helium nucleus: \(v_{He} = 0.8c\) Velocity of a lithium nucleus: \(v_{Li} = 0.5c\) Here, \(c\) represents the speed of light.
04

Calculate the de Broglie wavelength for each particle

Wavelength of an electron: \(\lambda_e = \frac{h}{m_e v_e} = \frac{h}{(9.11 \times 10^{-31} kg)(0.8c)}\) Wavelength of a proton: \(\lambda_p = \frac{h}{m_p v_p} = \frac{h}{(1.67 \times 10^{-27} kg)(0.2c)}\) Wavelength of a carbon nucleus: \(\lambda_C = \frac{h}{m_C v_C} = \frac{h}{(12 \times 1.67 \times 10^{-27} kg)(0.7c)}\) Wavelength of a helium nucleus: \(\lambda_{He} = \frac{h}{m_{He} v_{He}} = \frac{h}{(4 \times 1.67 \times 10^{-27} kg)(0.8c)}\) Wavelength of a lithium nucleus: \(\lambda_{Li} = \frac{h}{m_{Li} v_{Li}} = \frac{h}{(7 \times 1.67 \times 10^{-27} kg)(0.5c)}\)
05

Compare the de Broglie wavelengths

We can see that the denominators in each of the wavelengths determine their relative sizes. Since all velocities are fractional values of the speed of light, the smallest de Broglie wavelength will correspond to the particle with the highest mass and the highest velocity. Because a carbon nucleus has the highest mass and a velocity of \(70\%\) the speed of light, we expect that the de Broglie wavelength will be smaller than the other particle velocities. Therefore, the smallest de Broglie wavelength is from the carbon nucleus with a velocity of \(70\%\) the speed of light (option c).

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