- #1
saim_
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A beam of neutrons with energy E runs horizontally into a crystal. The crystal transmits half the neutrons and deflects the other half vertically upwards. After climbing to height H these neutrons are deflected through 90 degrees onto a horizontal path parallel to the originally transmitted beam. The two horizontal beams now move a distance [itex]L[/itex] down the laboratory, one distance [itex]H[/itex] above the other. After going distance [itex]L[/itex], the lower beam is deflected vertically upwards and is finally deflected into the path of the upper beam such that the two beams are co-spatial as they enter the detector. Given that particles in both the lower and upper beams are in states of well-defined momentum, show that the wavenumbers [itex]k[/itex], [itex]k^{′}[/itex] of the lower and upper beams are related by
[itex]{\large k \simeq k' \left( 1- \frac{m_{{\small N}}gH}{2E} \right) }[/itex]
Attempted Solution:
Let E be the kinetic energy of the neutrons in lower beam.
[itex]{\large k = \frac{\sqrt{2mE}}{h}}[/itex]
[itex]{\large k' \simeq \frac{\sqrt{2m(E - mgh)}}{h}}[/itex]
[itex]\Rightarrow {\large \frac{k}{k'} \simeq \sqrt{\frac{2mE}{2m(E-mgh)}} = \sqrt{\frac{E}{E-mgh}} }[/itex]
I have no idea how to go beyond this.
[itex]{\large k \simeq k' \left( 1- \frac{m_{{\small N}}gH}{2E} \right) }[/itex]
Attempted Solution:
Let E be the kinetic energy of the neutrons in lower beam.
[itex]{\large k = \frac{\sqrt{2mE}}{h}}[/itex]
[itex]{\large k' \simeq \frac{\sqrt{2m(E - mgh)}}{h}}[/itex]
[itex]\Rightarrow {\large \frac{k}{k'} \simeq \sqrt{\frac{2mE}{2m(E-mgh)}} = \sqrt{\frac{E}{E-mgh}} }[/itex]
I have no idea how to go beyond this.