1. h/√(3mkT) 2. 2h/√(3mkT) 3. 2h/√(mkT) 4. h/√(mkT)

Answer: 1. h/√(3mkT) Key idea: a thermal neutron has average kinetic energy 3/2kT , and λ=h/p=h/√(2mK) . λ=h/√(2m·3/2kT)=h/√(3mkT) .

The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is :

1. \(\frac{h}{\sqrt{3mkT}}\)

2. \(\frac{2h}{\sqrt{3mkT}}\)

3. \(\frac{2h}{\sqrt{mkT}}\)

4. \(\frac{h}{\sqrt{mkT}}\)

Doubtly Asked 30 Oct 2026 NEET UG 2017 Practice test neet Save WhatsApp Report

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Answer: 1. \(\frac{h}{\sqrt{3mkT}}\)

Key idea: a thermal neutron has average kinetic energy \(\tfrac32kT\), and \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mK}}\).

\(\lambda=\dfrac{h}{\sqrt{2m\cdot\frac32kT}}=\dfrac{h}{\sqrt{3mkT}}\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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