Question

Derive the approximate form of Heisenberg’s uncertainty principle for energy and time, ΔEΔt≈h, using the following arguments: Since the position of a particle is uncertain by Δx≈λ, where λ is the wavelength of the photon used to examine it, there is an uncertainty in the time the photon takes to traverse Δx. Furthermore, the photon has an energy related to its wavelength, and it can transfer some or all of this energy to the object being examined. Thus the uncertainty in the energy of the object is also related to λ. Find Δt and ΔE; then multiply them to give the approximate uncertainty principle.

a) Δt = h/Δx, ΔE = h/λ
b) Δt = h/λ, ΔE = h/Δx
c) Δt = λ/h, ΔE = Δx/h
d) Δt = Δx/h, ΔE = λ/h

Answer 1

The correct derivation yields (b) Δt = h/λ, ΔE = h/Δx as the approximate form of Heisenberg's uncertainty principle for energy and time.

Step-by-step explanation:

Heisenberg's uncertainty principle states that the product of the uncertainties in position (Δx) and momentum (Δp) of a particle is greater than or equal to Planck's constant divided by 2 (h/2π): ΔxΔp ≥ h/2π. To relate this to energy and time uncertainties, we can use the de Broglie wavelength, λ = h/p, where p is the momentum. Since Δx is approximately equal to the wavelength (Δx ≈ λ), we can express the uncertainty in momentum as Δp ≈ h/Δx.

Now, consider the relationship between energy (E) and momentum for a photon: E = pc, where c is the speed of light. We can rewrite this as ΔE ≈ Δp c. Substituting Δp ≈ h/Δx, we get ΔE ≈ (h/Δx) c. Now, since c/Δx is the time it takes for the photon to traverse Δx, we have ΔEΔt ≈ h, where Δt = Δx/c.

Therefore, the correct approximate form of Heisenberg's uncertainty principle for energy and time is Δt = h/λ, ΔE = h/Δx, corresponding to option (b). This demonstrates the interconnectedness of position uncertainty, time uncertainty, and the inherent limits in measuring both energy and time simultaneously, as described by Heisenberg's uncertainty principle.