Compare the de Broglie wavelength of an alpha particle moving at 3.40 times 10^7 miles per hour (1.52 times 10^7 mathrm(m) / mathrm(s) ) to that of a baseball moving at 90.0 miles per hour (40.2 mathrm(~m) / mathrm(s)) and an electron with a speed of 1.30 times 10^7 miles per hour (5.81 times 10^6 mathrm(~m) / mathrm(s)) . Particle & Mass (kg) & }(c) Velocity (mathrm(m) / mathrm(s)) & Wavelength & Region alpha particle & & baseball & 0.140 & 40.2 & 1.175 times 10^wedge-34 & smaller than 10^-20 mathrm(~m) electron & 9.11 times 10^-31 & 5.81 times 10^6 & &

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Compare the de Broglie wavelength of an alpha particle moving at 3.40 times 10^7 miles per hour (1.52 times 10^7 mathrm(m) / mathrm(s) ) to that of a baseball moving at 90.0 miles per hour (40.2 mathrm(~m) / mathrm(s)) and an electron with a speed of 1.30 times 10^7 miles per hour (5.81 times 10^6 mathrm(~m) / mathrm(s)) . Particle & Mass (kg) & }(c) Velocity (mathrm(m) / mathrm(s)) & Wavelength & Region alpha particle & & baseball & 0.140 & 40.2 & 1.175 times 10^wedge-34 & smaller than 10^-20 mathrm(~m) electron & 9.11 times 10^-31 & 5.81 times 10^6 & &

Answer 1

1. The de Broglie wavelength of an alpha particle is significantly smaller than that of a baseball. 2. The de Broglie wavelength of an electron is significantly larger than that of a baseball.

Answer 2

## Step 1: The de Broglie wavelength of a particle is given by the formula: ### \( \lambda = \frac{h}{p} \) where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} m^{2}kg/s \)) and \( p \) is the momentum of the particle. The momentum is calculated as the product of mass and velocity. ## Step 2: For an alpha particle with a mass of \( 6.64 \times 10^{-27} kg \) and a velocity of \( 1.52 \times 10^{7} m/s \), the de Broglie wavelength will be: ### \( \lambda_{\alpha} = \frac{h}{m_{\alpha} \times v_{\alpha}} \) ## Step 3: For a baseball with a mass of 0.140 kg and a velocity of 40.2 m/s, the de Broglie wavelength will be: ### \( \lambda_{baseball} = \frac{h}{m_{baseball} \times v_{baseball}} \) ## Step 4: For an electron with a mass of \( 9.11 \times 10^{-31} kg \) and a velocity of \( 5.81 \times 10^{6} m/s \), the de Broglie wavelength will be: ### \( \lambda_{electron} = \frac{h}{m_{electron} \times v_{electron}} \) ## Step 5: The de Broglie wavelength of the baseball is given as \( 1.175 \times 10^{-34} m \). We can use this as a reference to compare the calculated wavelengths. ## Step 6: The de Broglie wavelength of a particle is inversely proportional to its mass and velocity. Therefore, the larger the mass or the greater the velocity of the particle, the shorter the de Broglie wavelength.

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