The de-Broglie wavelength of neutron at 127°C is:
(Given Boltzmann constant, $k=1.38 × 10^{-23} J\, mole^{-1}K^{-1}, h=6.63 × 10^{-34} Js$, mass of neutron = $1.66 × 10^{-27} kg$)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.264 Å
The De-Broglie Wavelength (λ) of the neutron is,
$λ=\frac{h}{P}=\frac{h}{m_nv}=\frac{h}{\sqrt{2m_ek}}$
where,
$P=m_nv$, Momentum of neutron
$m_n$, Mass of neutron
v, velocity of neutron
k, kinetic energy of neutron
$∴λ=\frac{6.63 × 10^{-34}}{\sqrt{2×1.66×10^{-27}×1.36×10^{-23}}}$
$=\frac{6.63}{\sqrt{2×1.66×1.38}}×10^{-10}$
$=1.26×10^{-10}m$