An 81 -kg student who has just studied matter waves is concerned that he may be diffracted as he walks through a doorway that is \(81 \mathrm{cm}\) across and \(12 \mathrm{cm}\) thick. (a) If the wavelength of the student must be about the same size as the doorway to exhibit diffraction, what is the fastest the student can walk through the doorway to exhibit diffraction? (b) At this speed, how long would it take the student to walk through the doorway?

Short Answer

Expert verified
Answer: The maximum speed at which a student can walk through the doorway to exhibit diffraction is approximately 1.01 x 10^(-36) m/s. To walk through the doorway at this speed, it would take the student approximately 1.19 x 10^(38) seconds.

Step by step solution

01

Calculate the student's wavelength using de Broglie equation

The de Broglie equation relates the wavelength \(\lambda\) of a particle to its momentum \(p\) as follows: \(\lambda = \frac{h}{p}\) Where \(h\) is the Planck constant, \(p\) is the momentum of the particle which can be expressed as the product of the particle's mass \(m\) and its velocity \(v\). Therefore, we have the following equation: \(\lambda = \frac{h}{m v}\)
02

Find the maximum speed of the student

In this case, the wavelength of the student must be approximately equal to the width of the doorway (81 cm) to exhibit diffraction. Now, we know the mass of the student (81 kg) and the Planck constant (\(6.626 \times 10^{-34} \mathrm{Js}\)). We need to find the student's speed (\(v\)) using the following equation: \(\lambda = \frac{h}{m v}\) We can rearrange the equation above to solve for the student's speed: \(v = \frac{h}{m \lambda}\) Plugging in the values, we get: \(v = \frac{6.626 \times 10^{-34} \mathrm{Js}}{81 \mathrm{kg} \times 0.81 \mathrm{m}}\) \(v \approx 1.01 \times 10^{-36} \mathrm{ms^{-1}}\) So, the maximum speed a student can walk through the doorway to exhibit diffractive effects is approximately \(1.01 \times 10^{-36} \mathrm{ms^{-1}}\).
03

Calculate the time taken to walk through the doorway

To find the time taken for the student to walk through the doorway at this speed, we can use the simple formula for speed: \(v = \frac{d}{t}\), where \(d\) is the distance traveled (12 cm in this case) and \(t\) is the time taken. Rearranging the formula, we get: \(t = \frac{d}{v}\) Plugging in the values, we get: \(t = \frac{0.12 \mathrm{m}}{1.01 \times 10^{-36} \mathrm{ms^{-1}}}\) \(t \approx 1.19 \times 10^{38} \mathrm{ s}\) So, it would take the student approximately \(1.19 \times 10^{38}\) seconds to walk through the doorway at the speed where diffraction effects may be observed.

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