CUET · PHYSICS · PYQ PAPER 2023
The de-Broglie wavelength of neutron at \( 127^\circ \text{C} \) is : (Given Boltzmann constant, \( k = 1.38 \times 10^{-23} \text{ J mole}^{-1} \text{K}^{-1}, h = 6.63 \times 10^{-34} \text{ Js} \), mass of neutron \( m = 1.66 \times 10^{-27} \text{ kg} \))
- A \(1.264 \AA\)
- B \(12.64 \AA\)
- C \(8.28 \AA\)
- D \(828 \AA\)
Answer & Solution
Correct Answer
(A) \(1.264 \AA\)
Step-by-step Solution
Detailed explanation
\( T = 127 + 273 = 400 \text{ K} \) \( \lambda = \frac{h}{\sqrt{3mk_BT}} \) \( \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 1.66 \times 10^{-27} \times 1.38 \times 10^{-23} \times 400}} \) \( \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2.74896 \times 10^{-47}}} \)…
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