Question 16. At what speed is an electron's de Broglie wavelength (a) 1.0 pm, (b) 1.0 nm, (c) 1.0 um, and (d) 1.0 mm?

Step 1Subpart (a):Note: Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered.According to de Broglie, every particle has a wave aspect. The wave associated with any particle is called as matter wave.Step 2The formula to calculate the de Broglie wavelength of the electron is given byλ=hmv....................(1)Here, λ is the de Broglie wavelength, h is Planck's constant, m is the mass of the electron and v is its speed.Step 3Substitute 1.0 pm for λ, 6.626×10-34 J·s for h and 9.1×10-31 kg for m into equation (1) and solve to calculate the speed of the electron.1.0 pm=6.626×10-34 J·s9.1×10-31 kg×vv=6.626×10-34 J·s9.1×10-31 kg×1.0 pm×1 m1012 pm≈7.3×108 m/sStep 4Answer: The required speed of the electron is 7.3×108 m/s.Step 5Subpart (b):Substitute 1.0 nm for λ, 6.626×10-34 J·s for h and 9.1×10-31 kg for m into equation (1) and solve to calculate the speed of the electron.1.0 nm=6.626×10-34 J·s9.1×10-31 kg×v v=6.626×10-34 J·s9.1×10-31 kg×1.0 nm×1 m109 nm ≈7.3×105 m/sStep 6Answer: The required speed of the electron is 7.3×105 m/s.Step 7Subpart (c):Substitute 1.0 μm for λ, 6.626×10-34 J·s for h and 9.1×10-31 kg for m into equation (1) and solve to calculate the speed of the electron.1.0 μm=6.626×10-34 J·s9.1×10-31 kg×v v=6.626×10-34 J·s9.1×10-31 kg×1.0 μm×1 m106 μm ≈7.3×102 m/sStep 8Answer: The required speed of the electron is 7.3×102 m/s. ...