De Broglie Wavelength Of A Thermal Neutron
Inside a reactor a neutron of mass 1.67×10−27 kg has been slowed to thermal equilibrium with surroundings at a temperature of 300 K. What is the de Broglie wavelength associated with such a thermal neutron?
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Solution
1.45 A˚
A particle in thermal equilibrium has average translational kinetic energy K=23kBT, so its momentum is p=2mK=3mkBT and its de Broglie wavelength is λ=3mkBTh. Substituting m=1.67×10−27 kg, kB=1.38×10−23 J K−1 and T=300 K, the quantity under the root is 3×1.67×10−27×1.38×10−23×300≈2.07×10−47, whose square root is about 4.55×10−24 kg m s−1. Then λ=4.55×10−246.6×10−34≈1.45×10−10 m=1.45 A˚. The value 0.29 A˚ wrongly inflates the temperature or momentum. The value 14.5 A˚ misplaces a power of ten. The value 0.91 A˚ omits the factor of three from the kinetic-energy expression. Crucially, this wavelength is comparable to interatomic spacings, which is precisely why thermal neutrons are superb probes for crystal-structure determination by neutron diffraction, building on the same de Broglie idea highlighted in NCERT. A magnitude check confirms a near-angstrom value, consistent with the slow speeds of thermalised neutrons. A subtle point is that the average thermal energy used here is a statistical mean, so individual neutrons have a spread of speeds and therefore a distribution of wavelengths about this central value. Nonetheless, because the typical wavelength lands near one angstrom, beams of thermal neutrons rival X-rays for probing crystal structures, with the added advantage of being scattered strongly by light atoms such as hydrogen.
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About This Question
- Subject
- physics
- Chapter
- dual nature of radiation and matter
- Topic
- de broglie wavelength of a thermal neutron
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
1.45 A˚
A particle in thermal equilibrium has average translational kinetic energy K=23kBT, so its momentum is p=2mK=3mkBT and its de Broglie wavelength is λ=3mkBTh. Substituting m=1.67×10−27 kg, kB=1.38×10−23 J K−1 and T=300 K, the quantity under the root is 3×1.67×10−27×1.38×10−23×300≈2.07×10−47, whose square root is about 4.55×10−24 kg m s−1. Then λ=4.55×10−246.6×10−34≈1.45×10−10 m=1.45 A˚. The value 0.29 A˚ wrongly inflates the temperature or momentum. The value 14.5 A˚ misplaces a power of ten. The value 0.91 A˚ omits the factor of three from the kinetic-energy expression. Crucially, this wavelength is comparable to interatomic spacings, which is precisely why thermal neutrons are superb probes for crystal-structure determination by neutron diffraction, building on the same de Broglie idea highlighted in NCERT. A magnitude check confirms a near-angstrom value, consistent with the slow speeds of thermalised neutrons. A subtle point is that the average thermal energy used here is a statistical mean, so individual neutrons have a spread of speeds and therefore a distribution of wavelengths about this central value. Nonetheless, because the typical wavelength lands near one angstrom, beams of thermal neutrons rival X-rays for probing crystal structures, with the added advantage of being scattered strongly by light atoms such as hydrogen.
This hard difficulty physics question is from the chapter dual nature of radiation and matter, covering the topic of de broglie wavelength of a thermal neutron. It appeared in the 2025 exam.
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